(SR) Lorentz t', x' = Intervals
Eleaticus
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(SR) Lorentz t', x' = Intervals
(c) Eleaticus/Oren C. Webster
Thnk...@concentric.net
------------------------------
Subject: 1. Introduction with the obvious debunking
of the use of 'just coordinates' in any
scientific formula.
Defenders of the Special Relativity faith are especially
fond of telling opponents of their space-time fairy tales
that they do not know the difference between coordinates
and magnitudes.
That may often be so, but the fault lies ultimately with
SR dogma. The Lorentz-Einstein transformations cannot
possibly be 'just coordinates', which is the interpre-
tation required to support the many sideshow carnival acts
with which the SR faithful bedazzle the public, and establish
their moral and intellectual superiority.
If I get in my car and drive steadily for a few hours at 50
kilometers per hour, is 50t the distance I travel?
Of course not. The last time my hours-counting 'just coord-
inates' clock was set to zero was when Zeno first reported
one of his paradoxes to Parmenides.
That was a long time ago, so my t is not useful for such
purposes unless you also use my clock to established the starting
time, perhaps t0, and use the formula 50(t-t0) to calculate the
distance.
In any case, my t is even then not 'just a coordinate' because
it always represents particular elapsed times that can be
used in the (t-t0) form to calculate perfectly good time
intervals (elapsed times).
Alternatively, I could (re)set my clock to zero at the start
of some meaningful time interval, in which case my t shows a
scientifically perfect current and/or end time.
In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a function
of an elapsed time interval (not 'just a coordinate') and a time
interval (-vx/cc; the interval amount the t' clock is being
screwed up at time t) and thus cannot be 'just a coordinate'
since neither of the independent variables is such a 'just' thing.
{Their meaning is shown below, step-by-step.]
If it takes me 50 minutes to cross the Interstate highway,
was x/50 my velocity crossing it?
Of course not. The origin of all my axes is at the very
spot where Zeno first presented his first paradox to
Parmenides. That makes my x equal a couple of thousands of
miles, plus, and is not useful for such purposes unless
you establish the starting x value, perhaps x0, and use the
formula (x-x0)/50 to calculate my velocity.
In any case, even then my x is not 'just a coordinate'
because it always repesents particular distance intervals
that can always be used in the (x-x0) form for any and every
scientific purose.
Alternatively, I could move my x-axis origin to the starting
(zero) point of some meaningful distance, in which case my x
shows a scientifically perfect current and/or end distance.
In which case, the Lorentz-Einstein x'=(x-vt)/g is a function
of a current/ending distance interval (not 'just a coordinate')
and a distance interval (-vt; the interval amount the x' axis
is being screwed up at time t) and thus cannot be 'just a coordinate'
since neither of the independent variables is such a 'just' thing.
{Their meaning is shown below, step-by-step.]
------------------------------
Subject: 2. Table of Contents
1. Introduction with the obvious debunking
of the use of 'just coordinates' in any
scientific formula.
2. Table of Contents.
3. The Lorentz-Einstein transforms.
4. The 'just coordinates' argument.
5. Single-system, little-purpose ambiguity.
6. Relating two coordinate measures/systems.
7. Distances and moving coordinate axes.
8. Time intervals.
9. Einstein's (1905) derivations.
10. A word about intervals.
11. Intervals versus the Twins Paradox.
12. Summary
------------------------------
Subject: 3. The Lorentz-Einstein transforms
Special Relativity's space-time circus is based on
the 'transformation' equations by which it is believed
one can relate a nominally 'stationary' system's space
and time coordinates to those of an inertially (not
accelerating) moving other observer.
That moving observer's own physical body and coordinate
system might have been identical in size to those of the
stationary observer before the traveller began moving,
but are 'seen' as very different by the stationary observer
when the relative velocity of the two is great enough, a
high percentage of the velocity of light.
Concerning ourselves - as is customary - with just
the spatial coordinate axis that lies parallel to
the direction of motion, and with time, Einstein
arrived at these formulas that relate the moving
system measures or coordinates (x' and t') to the
stationary system coordinates (x and t):
x' = (x - vt)/sqrt(1-vv/cc) (Eq 1x)
t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)
The v is for the two systems' relative velocity as seen
by the stationary observer, and is positive if the dir-
ection is toward higher values of x. By concensus,
the moving system x'-axis higher values also lie in
that direction, and all axes parallel the other system's
corresponding axis.
We used vv to mean the square of v but might use v^2
for that purpose below. Similarly for c.
Because it is believed that no physical object can
reach or exceed c, the square-root term in both
denominators is presumed always less than one, which
means that the formulas say both x' and t' will tend to
be greater than x and t, respectively. However,
SRians call the x' result 'contraction' - which means
shortening - and the t' result 'dilation' - which
means increasing.
------------------------------
Subject: 4. The 'just coordinates' argument
The 'just coordinates' argument is so patently ridiculous
that even opponents have a hard time accepting just how
simple and obvious its debunking can be, as shown in this
section. However, further sections take a more arithmet-
ical approach that you'll maybe find more professorial.
The 'just coordinates' argument is that t is mot a
duration, not a time interval; it's just an arbitrary
clock reading. But what if the moving system observer
comes speeding by while you make your annual 'spring
forward' or 'fall back' change? The formula says that
the moving system clock's 'just coordinate' reading
can be calculated from yours:
t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)
Imagine the moving system oberver's confusion if his
clock changes its reading while he's looking at it!
If his clock doesn't change when yours does, the formula
is wrong; if it is truly a 'just coordinates' formula.
And then what happens if you realize you were a day
early and put your clock back to what it had said
previously?
And suppose you are in NYC and your twin in LA and
both are watching the moving observer. You'll both be
using the same v because you are at rest wrt (with
respect to) each other. You're on Eastern Standard
Time and your twin is on Pacific Standard Time
maybe. You have three hours more on your clock than
does your twin.
On which 'just coordinate' clock will the Lorentz
transforms base the 'just coordinate' time the moving
system clock says? The formula applies to both of
your t-times:
t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)
Sure, the idea that you can change someone else's
clock with no connection of any kind is really
ridiculous, but Eqs 1x and 1t aren't MY equations.
Are they yours? And we aren't the ones to say x, t,
x', and t' are just coordinates.
If the t' formula is actually either an elapsed
time formula, or the basis of a t'/t ratio, then
there is no implication that one clock's reading
has anything to do with the other's.
It can only be rates of clock ticking, or how one
time INTERVAL compares to the other that the formula
is about.
------------------------------
Subject: 5. Single-system, little-purpose ambiguity.
Since we're going to be comparing measurements on two
coordinate systems in the next section, let's go to
our supply cabinet and get our yard-stick (which we
use to measure things in inches) and our meter-stick
(which we use to measure things in centimeters).
Here, I'm getting mine. Oh! Oh!
There's an ant on mine, and he ... she ... sure is
hanging on, right at the 3.5 inch mark of the yard-
stick.
Let's see if I can wave the stick around enough that
she'll let go. Nope.
However, before I gave up I waved the stick and the
ant 'all over the place".
Always, however, the ant was at the 3.5" mark on the
yard-stick, and always 3.5" away from the end of the
stick, however far and wide I have transported her.
Neither of those 3.5" facts means very much. Of the
two, the distance aspect meant almost nothing. So
the distance was 3.5" from the end. So what? That
length, distance, was not in use. And only maybe
the ant might have been concerned with just what
location, 'just coordinate', on the stick she was
at.
Just so with x and t.
So, is the 3.5" reading just a coordinate? Or a
distance/length? It's ambiguous in and of itself,
and really makes no difference what you say until
you try to make use of the number.
Hey, my address is 5047 Newton Street. If you
are looking for me and you're at 4120 Newton, it
is helpful information, because it tells you which
direction to go. Is that 'just coordinate'?
Where it really becomes useful, perhaps, is in
telling you how far away I am. That's not just
a coordinate value, that's a distance, length,
interval.
However, it is subtracting 4120 from 5047 that
tells you which direction and how far. It is only
because both 5047 and 4120 are distances from the
same point - ANY same point - that the result means
anything.
My x - my yardstick reading - is always a distance
or length; it is impossible to be otherwise with
an honest, competently designed yardstick.
Whether or not its reading is of good use in some
particular scientific formula depends on whether
I put the zero end of the yardstick at some useful
place. As in the introduction, we should either
put it at the starting location/end, or use two
readings from it: (x-x0).
------------------------------
Subject: 6. Relating two coordinate measures/systems.
Taking care to not damage our brave little ant, I place
my yard-stick onto the table, zero end to the left, 36"
end to the right.
Now I place the 'just coordinate' meter-stick on the table
in the same orientation, in a random location, and find
that the ant's coordinate on the meter-stick is 51.
The formula relating centimeters to inches is cm=i*2.54
but we want a formula similar to x'=(x-vt)/sqrt(1-vv/cc).
That would be c=i/.03937 approximately, but let's use x'
for the meter-stick reading, and x for the inch reading:
x'=x/.3937.
3.5/.3937 = 8.89
Wait a minute. It's not just science but definition
that says c=i/.3937=8.89, so something is wrong. 8.89
is not 51.
We already knew that 51 cm was just an arbitrary coordinate.
Arbitrary not because that point isn't 51 cm from the zero
end of the meter-stick, but because the zero point was in an
arbitrary position.
Let's put the meter-stick in a position where it's
zero point is at the yard-stick zero point.
What is the centimeter coordinate now? Hey. 8.89,
just like the formula says.
The only way for a 'transform' like x'=x/g to work,
whatever g might be, is for both coordinate systems
to have their zero points aligned, in which case
saying the two measures are not intervals is pure
idiocy.
Noe that with both zero points at the same position
both x' and x are great measures for scientific
purposes, in any and every case where we were smart
enough to put those zero points at a useful location.
There is one extension of x'=x/g that will let us
use the meter-stick in arbitrary position.
When the cm reading was 51, the zero point of the
yard-stick read (51-8.89=) 42.11 cm. If we call that
point x.z' we get
x' = x.z' + x/.3937.
= 42.11 + 3.5/.3937
= 42.11 + 8.89
= 51.
Obviously, in this formula x/.3937 is the distance
from the x' coordinate of the location where x=0.
An interval.
Just as obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate, does not mean that x'
is anything more than nonsense for use in any
scientific formula.
Unless we were smart enough to put the x zero
point in a useful location, and use (x'-x.z') in
the scientific formula. (x'-x.z') equals the useful,
Ratio Scale value x/.3937.
So, we have discovered a basic fact: a transformation
formula like x'=x/g works only if the two zero points
of the coordinate systems coincide. That makes it non-
sense to say the two coodinates are only coordinates
and not intervals. Both must be values that represent
distances from their respective zero points unless you
take the proper steps to adjust for the discrepancy.
Make sure you understand that although the inclusion
of x.z' made it possible to correctly calculate x',
the result is nonsense when it comes to use of x'
for general length/distance purposes; it is x'-x.z'
that is a useful number in such cases. It could be
that we're measuring a sheet of paper with one end
at x=0 and the other at x=3.5; x'=51 is nonsense as
a centimeter measure of the paper.
But, you say, the Lorentz transform contain a -vt term.
------------------------------
Subject: 7. Distances and moving coordinate axes.
We discovered x'=x.z' + x/g as the correct formula
for relating one coordinate to another system's.
But the Lorentz transform contains another term,
-vt/sqrt(1-vv/cc). What is it?
Let's start with our x'=51 cm, x=3.5", x.z'=42.11 example.
Every minute, let's move the meter-stick one inch to our
right.
At minute 0, the cm reading was 51 cm.
At minute 1, the cm reading is now 50 cm.
At minute 2, the cm reading is now 49 cm.
In this instance, v=1 inch/minute. And t was 0, 1, 2.
What has happened is that we have made our x.z' a lie,
and increasingly so. -vt/.3937 is the change in x.z'.
x' = (x.z - vt/.3937) + x/.3937.
Obviously, vt/.3937 is not a coordinate; even most SRians
wouldn't imagine it was. It is an interval, the distance
over which the moving system has moved since t=0.
And, of course, x/.3937 is the distance of our brave
little ant from the point where x=0 and the centimeter
reading is x.z'-vt/.3937. Yes, every minute the meter-
stick moves to the right and the meter-stick coordinate
of the spot where x=0 gets less and less - and eventually
negative.
Make sure you understand that every minute the x'
coordinate, because of -vt/g, becomes a better measure
of, say, the 3.5" paper we might be measuring with
the yard-stick, given that 51 was too big a number and
-vt is negative. That is, until the two origins coincide
at x'=x=0, and then it gets worse and worse.
With -vt positive (because v<0) the situation is different.
With 51 and -vt positive, x' just gets worse and worse
over time.
Quite obviously, the fact that we now have the
correct formula for relating an x interval to an
arbitrary x' coordinate even when the x' axis is
moving, does not mean that x'is anything more than
nonsense for use in any scientific formula.
Unless we were smart enough to put the x zero point
in a useful location, and use (x'-x.z'+vt/.3937) in
the scientific formula. (x'-x.z'+vt/.3937) equals the
useful, Ratio Scale value x/.3937.
------------------------------
Subject: 8. Time intervals.
Instead of using our sticks, let's get out two clocks.
Mind you, we're not going to deal with different clock
rates here, just establish the same basics as for distance.
Your clock says 9:00 Eastern Standard Time (EST) and we
note that t=540 minutes when we put down the clock.
Blindly, let's turn the setting knob of your twin's Pacific
Standard Time clock and put it down before us.
According to what we see, EST's 540 minutes (9:00) corre-
sponds to PST's 14:30; t'=870.
We know the formula relating PST to EST is t' (pacific)
= t (eastern) - 180 (minutes). Thus, it is not correct
that the second clock can have an arbitrary setting,
because 870 <> 540-180.
We know that the two clocks are related by t' = t/1 since
both are using the same second, hour, etc units. But 870
(14:30 in minutes) is not 540/1-180, so once again we know
something is wrong.
However, t'=t.z' + t/1 works. EST midnight equals PST 0.0
(midnite) - 180, so t.z' = -180, and
t' = -180 + 540/1 = 360.
Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes.
We see thus that like distance measures/coordinates, time
axis origins (zero points) must either be 'lined up' or
adjusted for.
So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the moving
system elapsed time interval since the time axes were both at
a common zero. There is no t.z' adjustment:
t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)
Make sure you understand that in the clock case, if the
EST is showing a good number for elapsed time since the
travelling observer passed NYC, then the PST clock is
silliness. t.z' must be zero or must be taken out of
time lapse calculations for the PST clock to be used
intelligently, just as was true for x.z'.
What is lacking as yet for Lorentz t' is the -vx/cc term that
corresponds to the x' formula -vt term.
Break it up into two parts: v/c and x/c.
v/c is a scaling factor that changes velocity from whatever
kind of unit you are using over to fractions of c.
x/c is distance divided by velocity, which is time. x/c
is thus the time interval since the two time axes
had a common zero point - which they have to have in the
Lorentz transforms which do not have the t.z' term we
learned to use above.
Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the
moving system clock has been changed - since the common/
adjusted time - over and beyond the elapsed time interval
represented by x/sqrt(1-vv/cc).
We have discovered that the only way for t' to be t/g
is for t' and t to have a common zero point, just as
for x' and x. It would be otherwise if the t' formula
contained an adjustment t.z' under some name or other,
but the necessity to include such a term correlates
100% with t' numbers that aren't directly usable.
As for x and x', our knowledge of how to setup a proper
formula relating t and t' is of no use unless we use
the knowledge in scientific formulas; (t'-t.z'+xv/gcc)
gives us the only directly useful value: t/g.
------------------------------
Subject: 9. Einstein's (1905) derivations.
When we return to Einstein's derivations of the transform
formulas with a well-focused eye, we find he was a wee bit
confused - or at least self-contradictory.
When he set up his (at first unknown) tau=moving system
time formulas, he created three particular instances of tau.
Tau.0 is the time at which light is emitted at the moving
origin toward a mirror to the right that is moving at rest
wrt that moving origin and at a constant distance from that
origin. He lets the stationary time slot have the value t,
a constant, the stationary system starting time.
Tau.1 is the time at which the light is reflected. He
lets the stationary time be t+x'/(c-v); t is still a
constant and x'/(c-v) is the time interval since t.
Tau.2 is the time at which the light gets (back) to the
moving origin. The stationary time value is put as t +
x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v)
+ x'/(c+v) is the time interval since t.
On the thesis that the moving observer sees the time to
the mirror as the same as the time back to the origin,
he sets
.5[ tau.0 + tau.2 ] = tau.1.
Tau.0 completely drops out of the analysis and leaves
no trace, and has no effect.
Further, the t you see in tau.0, tau.1, and tau.2 also
completely drops out with no trace and no effect, leaving
us with exactly what you'd get if you had explicilty said
t' is an interval and so is t.
What doesn't drop out in the stationary time values is
x'/(c-v) and x'/(c+v), the time interval it takes for
light to get to the fleeing mirror, and the time interval
it takes for light to get back to the approaching origin.
Thus, his resultant t' formula is strictly based on time
intervals in the stationary system. Time intervals since
some starting time, yes, but time intervals.
There is absolutely nothing in the derived formulas that
depends on arbitrary coordinates like the constant t in
the stationary time arguments.
Let's look at the x dimension; it is x'=x-vt [as x increases
by vt, the effect over time is x'=(x+vt)-vt)], which Einstein
explicitly sets up as a constant stationary distance.
He uses that x' not just in the time interval parts of the
stationary time arguments, but also in the x (distance)
stationary system argument for the tau at the time light
is reflected.
x' can't be the stationary system coordinate of the mirror
at that time. That value is x'+vt.
x' is explicitly an interval, distance.
Thus, the whole tau derivation of the t' formula is fully and
explicitly based on x' - a spatial length/distance/interval -
and the two time interals x'/(c-v) and x'/(c+v).
While we're at it, if the starting t is not zero, his
x'=x-vt formula is complete nonsense also. Given that
there was some L that was the mirror x-location and length
when the light is emitted, if t was already, say, 500, then
x'=L-vt could have been a very negative length.
------------------------------
Subject: 10. A word about intervals.
There are intervals, and there are intervals.
If we put our yard stick zero point at one end
of a piece of paper and read off the coordinate
at the other end of the paper, we have a good
measure of the paper's length, a Ratio Scale
measure. [Absolute temperature scales are ratio
scale.]
If instead we put the one end of the paper at the
one inch mark (or the zero end of the stick one
inch 'into' the length of the paper) we get measures
that are one inch off the true, ratio scale length.
The two messed up measures are still intervals,
but they are Interval Scale measures. [Household
temperature scales are interval scale, which is
why your physics and chemistry professors won't
let you use them without first converting to the
ratio scale absolute temperatures.)
t'=t/g and x'=x/g represent ratio scale measures,
given that t and x were ratio scalae to start with.
t'=t.z'+t/g and t'=t/g-vx/gcc are both interval
scale measures, even given a good ratio scale t
and a good ratio scale x.
x'=x.z'+x/g and x'=x/g-vt/g are both interval
scale measures, even given a good ratio scale x
and a good ratio scale t.
Look for the "(SR) Lorentz t', x' = degraded measures"
document soon at a newsgroup near you.
------------------------------
Subject: 11. Intervals versus the Twins Paradox.
t'=(t-vx/cc)/g shows t' being greater than t.
The reason Special Relativity will not allow the
use of its basic time equation in determining what
SR has to say about the twins' ages, is that t' and
x' are supposedly just coordinates, and they say you
have to take the coordinate pairs (t',x') and (x,t)
into consideration in both the time and place the
twins' separation started and the time and place the
twins reunited.
Since t' and x' are actually both intervals, not
just coordinates, the 'excuse' is spurious, and is
so even without use of the obvious (x_b-x_a) and
(t_b-t_a) usages.
However, SR is right to be embarrassed by their
transformation formulas.
Look for the "(SR) Lorentz t', x' = degraded measures"
document at a newsgroup near you.
------------------------------
Subject: 12. Summary
A. t'=t/g and x'=x/g can be almost 'just coordinates'
in the sense that the values obtained may not be
of much use except in the most primal and useless
way: how long and how far since/from the time/
place they were zero. Even here, however, the zero
points within each of the two scale pairs (t',t)
and (x'.x) must have been lined up. If the zero
points have been intelligently selected (such as
at the starting point and time of a trip) they
can be rationally used 'as is' in any valid sci-
entific equation.
B. Even the interval scale t'=t.z' - xv/gcc + t/g and
x'=x.z' - vt/g + x/g are not 'just coordinates'. They
can be used to good effect by establishing the relevant
starting times/points and using (t'-t.z'+xv/gcc) and
(x'-x.z'+vt/g), as the situation may require.
C. When you see vx/gcc or vt/g in use in any guise with non-zero
values, you know the resultant t' or x' is a degraded, interval
scale value.
E-X: Anytime you do not see what amounts to t.z' and xv/gcc in
the time case, or x.z' and vt/g in the distance case, you
know that the t' and/or x' in use are intervals. Period.
Y: Either set your clock to zero at the start of the relevant
time interval, or use (t-t0), with both being readings on
the same clock. Either move your x-axis origin to the starting
end or point, or use (x-x0), with both being readings on
the same axis.
Z: In _(SR) Lorentz t', x' = Degraded (Interval) Scales_ we see
that t' and x' satisfy the mathematical tests for/of interval
scales when -vt and -vx/cc are not zero; thus, they must
be intervals. When -vt and -vx/cc are zero, t' and x'
satisfy the much better mathematical definition of
ratio scales, and are thus not just mere intervals,
but (rescaled) good ones.
Eleaticus
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
! Eleaticus Oren C. Webster Thnk...@concentric.net ?
! "Anything and everything that requires or encourages systematic ?
! examination of premises, logic, and conclusions" ?
!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
Uncle Al
[snip lies]
[snip 1300 lines of trolled garbage]
eleaticus, Oren Webster, is a despised and stooopid troll,
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
"Several crimes against logic and science" Ha ha ha!
Originally trolled across sci.physics sci.physics.relativity
alt.physics sci.math sci.answers alt.answers news.answers
Psychotic ineducable boring troll Eleaticus,
Were there to be internal inconsistencies in SR (meaning
inconsistencies of a purely mathematical logical nature) that would
automatically lead to contradictions in number theory, itself, and
arithmetic, since the mathematics of Minkowski geometry is
equiconsistent with the theory of real numbers and with arithmetic.
Eleaticus explicitly demonstrates that he is completely ignorant of
multivariable calculus. He has no concept of the Chain Rule in
multivariable calculus. Consider his Galilean Transformation goo and
dribble:
t' = t,
x' = x - vt,
y' = y,
z' = z.
His refusal to accept that t' must be introduced as a separate
variable springs from a massive emprical stupidity re space and time
are described as a four-dimensional manifold, with four coordinates
instead of a time evolution of a three-dimensional manifold, and that
the change of coordinate system should be a change of four
coordinates, and not a time-dependent change of three coordinates.
This is particularly vital when it comes to fields over space and time
(electric and magnetic fields for example).
The transformation law for the differential operators under the
Galilean transformation is given by:
d/dt' = d/dt + v d/dx,
d/dx' = d/dx,
d/dy' = d/dy,
d/dz' = d/dz.
This shows the necessity of introducing a new variable t', since
partial differentiation with respect to t' (constant x', y', z') is a
different operation to partial differentiation with respect to t
(constant x, y, z). The above transformation law is determined by the
Chain Rule:
d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz,
d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz,
d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz,
d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz.
The presence of the term involving d/dx in the expression for d/dt' is
indicative of the fact that x depends on t' (x', y', z', being held
constant), as can be seen from the fact that the coefficient of d/dx
in the expression for d/dt' is dx/dt'. Because of the now
demonstrated fact that Eleaticus has no formal education in
multivariable calculus, he has managed, somehow, to get it into his
head that the presence of the term involving d/dx in the expression
for d/dt' is indicative of t' depending on x (t, y, z, being held
constant). Because of his stupidty Eleaticus cannot get the correct
transformation law for the differential operators under the Galilean
Transformation, and he cannot determine the invariance or otherwise of
Maxwell's Equations under the Galilean Transformation. The first
advice to Eleaticus is to learn multivariable calculus.
Eleaticus should not pretend that he can understand how to determine
invariance or otherwise of Maxwell's Equations under the Galilean
Transformation, or under the Lorentz Transformation, until he
understands the multivariable calculus which underlies such
considerations. Eleaticus is a loud idiot.
The homogeneous Maxwell equations are invariant under the Galilean
Transformation, with transformation laws:
E_x' = E_x,
E_y' = E_y - v B_z,
E_z' = E_z + v B_y,
B_x' = B_x,
B_y' = B_y,
B_z' = B_z.
The derivation of these transformation laws was determined using the
transformation laws for the differential operators given above. These
transformation laws have the additional advantage that they determine
the correct transformation for the force law, thus providing further
evidence in favour of the transformation law for the differential
operators, as above.
The inhomogeneous Maxwell equations are also invariant under the
Galilean transformation, with transformation laws:
E_x' = E_x,
E_y' = E_y,
E_z' = E_z,
B_x' = B_x,
B_y' = B_y + v/c^2 E_z,
B_z' = B_z - v/c^2 E_y,
\rho' = \rho,
J_x' = J_x - v \rho,
J_y' = J_y,
J_z' = J_z.
Note the the transformation laws for the charge density and current
density are as they should be under the Galilean transformation.
Homogeneous equations are invariant under the Galilean Transformation,
and inhomogeneous equations are invariant under the Galilean
Transformation, but Maxwell's Equations as a whole are NOT invariant
under the Galilean Transformation, since the transformation laws
required for the EM field for the two cases are inconsistent with each
other. The transformation law for the EM field which makes the
homogeneous equations invariant will not also make the inhomogeneous
equations invariant. The transformation law for the EM field which
makes the inhomogeneous equations invariant will not also make the
homogeneous equations invariant.
On the other hand, all of Maxwell's equations are invariant under the
Lorentz Transformation, with transformation laws:
E_x' = E_x,
E_y' = \gamma (E_y - v B_z),
E_z' = \gamma (E_z + v B_y),
B_x' = B_x,
B_y' = \gamma (B_y + v/c^2 E_z),
B_z' = \gamma (B_z - v/c^2 E_y),
\rho' = \gamma (\rho - v/c^2 J_x),
J_x' = \gamma (J_x - v \rho),
J_y' = J_y,
J_z' = J_z,
where \gamma = 1/sqrt(1 - v^2/c^2).
Idiot Oren Webster sees himself this way,
http://www.mazepath.com/uncleal/effete6.jpg
The entire remainder of the planet sees him this way,
http://www.mazepath.com/uncleal/effete3.png
<http://www.albinoblacksheep.com/flash/youare.swf>
http://www.mazepath.com/uncleal/sunshine.jpg
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html
<http://www.firehead.org/~jessh/film/kubrick/Kubrick-Psycho.html>
<http://www.naturalchild.com/elliott_barker/prisons.html>
Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24
GPS satellites carries either four cesium atomic clocks or three
rubidum atomic clocks in orbit, with full relativistic corrections
being applied.
Internal inconsistencies in SR (meaning inconsistencies of a purely
mathematical logical nature) automatically lead to contradictions in
number theory, itself, and arithmetic, since the mathematics of
Minkowski geometry is equiconsistent with the theory of real numbers
and with arithmetic.
<http://optoelectronics.perkinelmer.com/content/Datasheets/rfs2f.pdf>
<http://math.ucr.edu/home/baez/RelWWW/tests.html>
Mathematics of gravitation
<http://wugrav.wustl.edu/people/CMW/update98.pdf>
<http://www.astro.northwestern.edu/AspenW04/Papers/lorimer1.pdf>
Equivalence Principle testing
http://arXiv.org/abs/hep-th/0111236
Geometric structure of reality
http://arxiv.org/abs/gr-qc/0103044
http://arXiv.org/abs/hep-th/0307140
GR structure, especially Part 4/p. 7
<http://relativity.livingreviews.org/Articles/lrr-2001-4/index.html>
http://arXiv.org/abs/gr-qc/0311039
<http://www.weburbia.demon.co.uk/physics/experiments.html>
Experimental constraints on General Relativity
<http://tycho.usno.navy.mil/ptti/ptti2002/paper20.pdf>
Nature 425 374 (2003)
http://www.eftaylor.com/pub/projecta.pdf
<http://www.public.asu.edu/~rjjacob/Lecture16.pdf>
<http://relativity.livingreviews.org/Articles/lrr-2003-1/index.html>
Relativity in the GPS system
http://arXiv.org/abs/gr-qc/9909014
Amer. J. Phys. 71 770 (2003)
Phys. Rev. Lett. 92 121101 (2004)
falling light
<http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html>
<http://metrologyforum.tm.agilent.com/pdf/flying_clock_math.pdf>
http://metrologyforum.tm.agilent.com/cesium.shtml
http://arxiv.org/abs/physics/0008012
Hafele-Keating Experiment
http://www.hawaii.edu/suremath/SRtwinParadox.html
<http://physics.syr.edu/courses/modules/LIGHTCONE/twins.html>
Twin Paradox
Science 303(5661) 1143;1153 (2004)
http://arXiv.org/abs/astro-ph/0401086
http://arxiv.org/abs/astro-ph/0312071
<http://relativity.livingreviews.org/Articles/lrr-2003-5/index.html>
<http://skyandtelescope.com/news/article_1473_1.asp>
Deeply relativistic neutron star binaries
http://arxiv.org/abs/hep-th/0405160
Black hole evaporation
Physics Today 57(7) 40 (2004)
http://physicstoday.org/vol-57/iss-7/p40.shtml
No aether
http://fsweb.berry.edu/academic/mans/clane/
http://physicsweb.org/articles/world/17/3/7
No Lorentz violation
http://arXiv.org/abs/gr-qc/0409089
Spin-2 gravitons have problems
<http://groups-beta.google.com/group/sci.physics.strings/msg/ba31a00f5f26277a>
(so does the proposal)
http://arXiv.org/abs/gr-qc/0411113
<http://www.npl.washington.edu/eotwash/pdf/prl83-3585.pdf>
http://arXiv.org/abs/gr-qc/0301024
Phys. Rev. Lett. 93 261101 (2004)
Nordtvedt Effect
http://map.gsfc.nasa.gov/
http://arxiv.org/abs/astro-ph/0403292
http://arXiv.org/abs/astro-ph/0310723
WMAP + Sloane Digital Sky Survey
http://arxiv.org/abs/hep-ph/0404175
Dark matter candidates
<http://nedwww.ipac.caltech.edu/level5/March01/Carroll/frames.html>
Carroll on what it all means.
Special Relativity is physics on a topologically trivial Lorentzian
manifold with a metric whose curvature tensor is zero. This is a
perfectly diffeomorphism-invariant condition and does not require
any particular coordinate choice. It is invariant under
the full group of diffeomorphisms. The Poincare group is
the group of *isometries* of the metric in special relativity.
The Special Relativity metric is *non-dynamical* (unlike GR). It
defines the coupling *constants* of your theory. If you change the
metric in any nontrivial way you are changing your theory. An
operation can only be called a "symmetry" of a special-relativistic
(non-gravitational) theory if it preserves the metric, and therefore
the symmetry of special-relativistic theories is the Poincare group
only. General Relativity (gravitation) has a dynamic metric.
NIM A 355 537 (1995)
Physics Letters B 328 103 (1994)
Physical Review Letters 64 1697 (1990)
Physical Review Letters 39 1051 (1977)
Physical Review 135 B1071 (1964)
Physics Letters 12 260 (1964)
Europhysics Letters 56(2) 170-174 (2001)
General Relativity and Gravitation 34(9) 1371 (2002)
http://fourmilab.to/etexts/einstein/specrel/specrel.pdf
<http://www.geocities.com/physics_world/sr/ae_1905_error.htm>
<http://www.physics.gatech.edu/people/faculty/finkelstein/relativity.pdf>
Longitudinal and transverse mass
Physics Today 58(3) 34 (2005)
Time passage, equator vs. poles
http://arxiv.org/abs/gr-qc/0306076.pdf
<http://www.metaresearch.org/solar%20system/gps/absolute-gps-1meter-3.ASP>
http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf
http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm
http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm
http://www.trimble.com/gps/index.html
http://sirius.chinalake.navy.mil/satpred/
http://www.phys.lsu.edu/mog/mog9/node9.html
http://egtphysics.net/GPS/RelGPS.htm
http://www.schriever.af.mil/gps/Current/current.oa1
http://edu-observatory.org/gps/gps_books.html
<http://www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html>
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
eleaticus
"Uncle Al" <Uncl...@hate.spam.net> wrote in message
news:422B5855...@hate.spam.net...
> Eleaticus wrote:
> Psychotic ineducable boring troll Eleaticus,
> Eleaticus explicitly demonstrates that he is completely ignorant of
> multivariable calculus. He has no concept of the Chain Rule in
> multivariable calculus. Consider his Galilean Transformation goo and
> dribble:
> t' = t,
You have repeated that lie perhaps hundreds of times.
Just once show that there has been any time that I have asserted t'=t in the
context of Galilean transforms.
Liar! Liar! Diaper on fire!
Since it is not possible to meet that challenge, try this one: make just
one post that is both relevant and responsive in the terms posed by the
article you so despise.
eleaticus
The Ghost In The Machine
<elea...@bellsouth.net>
wrote
on Sun, 6 Mar 2005 14:20:36 -0600
<5GJWd.18785$%Y4....@bignews6.bellsouth.net>:
If one assumes, mathematically, that c' = c, then one has to
at least consider the possibility that t' != t.
Assume that (x,t) * M(v) = (x', t'), where M = [m00, m01, m10, m11]
is the usual 2x2 matrix-function transform (domain: R,
range: 2x2 matrices over R) with the following characteristics.
[1] (vt,t) * M(v) = (0,t') for some t'. (This is required because we're
specifying the second coordinate system moving with velocity v
relative to the first.)
[2] (0,t) * M(v) = (-vt',t') for some t'.
[3] (ct,t) * M(v) = (ct',t') for some t'. (OWLS invariance.)
[4] M(-v) * M(v) = I, the identity matrix [1 0 0 1].
[5] M(0) = I.
If we assume a priori that t' = t, then this places certain
constraints on the mxy(v), which are easy to compute; in
fact, (x,t) * M(v) = (x',t) requires that m01 = 0 and m11 = 1.
Since (0,t) * M(v) = (-vt',t') and t' = t, m10 = -v.
Since (vt,t) * M(v) = (0,t), ( (vt,t) - (0,t) ) * M(v) = (vt,0) * M(v)
= ( (0,t) - (-vt,t) ) = (vt,0), which means m00 = 1.
We have the Galilean transform M(v) = [1 0 -v 1]. Trouble is, we've
not looked at [3] yet:
(ct,t) * [1 0 -v 1] = ((c-v)t,t)
and we don't have OWLS invariance.
I won't work out here the goopy details why but the Lorentz
in this form can be written
L(v) = [g -vg/c^2 -vg g]
where g = 1/sqrt(1-v^2/c^2).
One can easily validate that this transform satisfies the requirements:
(vt,t) * L(v) = (gvt-gvt,-v^2gt/c^2+gt) = (0,t'). [1] down.
(0,t) * L(v) = (-gvt, gt) = (-vt',t'). [2] down. (Note that
we didn't make the requirement that [1] and [2] had to have equal t's!)
[g vg/c^2 vg g] * [g -vg/c^2 -vg g] = [g^2-v^2g^2/c^2 0 0 -v^2g^2/c^2+g^2];
since g^2 = 1/(1-v^2/c^2), g^2-v^2g^2/c^2 = 1. [4] down.
[5] is trivial as g(0) = 1.
It only remains to validate [3].
(ct,t)*[g -vg/c^2 -vg g] = (gct-gvt,-vgt/c+gt) = (gt(c-v), gt(c-v)/c).
I won't go into non-linear transforms since that's a bit out of
my competence range (and I'm not all that knowledgable about the
ins and outs of tensors), but I for one don't see them working either.
Besides, if one assumes SR the tensor must be linear anyway; otherwise
one has a curved space.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
eleaticus
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
message news:s5jsf2-...@sirius.athghost7038suus.net...
> If one assumes, mathematically, that c' = c, then one has to
> at least consider the possibility that t' != t.
OK assumption for the heck of it, but MMX fits c'=c+v.
> Assume that (x,t) * M(v) = (x', t'), where M = [m00, m01, m10, m11]
> is the usual 2x2 matrix-function transform (domain: R,
> range: 2x2 matrices over R) with the following characteristics.
> [1] (vt,t) * M(v) = (0,t') for some t'. (This is required because we're
> specifying the second coordinate system moving with velocity v
> relative to the first.)
Well, t' is not really a requirement in the context of the thread's
immediate material, where I point out Uncle assAl lies a lot.
But, maybe no harm done. We'll see.
> [2] (0,t) * M(v) = (-vt',t') for some t'.
That is just flat, error in the context you presented.
Correct is
[2] (0,t) * (M(v) = (-vt, t') for some t'.
If t'<>t the one thing for sure is that vt<>vt' and -vt is the transform of
0.
> [3] (ct,t) * M(v) = (ct',t') for some t'. (OWLS invariance.)
That, too, is just flat error in the context you presented.
Correct is
[3] (ct,t) * M(v) = (ct-vt,t') for some t'.
ct gives us an x-value which transforms as ct-vt, and with t<>t ct<>ct'.
We need to start over. Way too much error for the rest to be meaningful.
Thanks for the go at it!
eleaticus
The Ghost In The Machine
<xgPWd.3988$c72....@bignews3.bellsouth.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:s5jsf2-...@sirius.athghost7038suus.net...
>> If one assumes, mathematically, that c' = c, then one has to
>> at least consider the possibility that t' != t.
>
> OK assumption for the heck of it, but MMX fits c'=c+v.
Without a moving light source MMX can't say either way;
emission theory and SR predict exactly the same thing.
The only thing MMX *disproves* is the absolute aether theory.
(That, and giving a very precise [for its time] lightspeed measurement.)
>
>> Assume that (x,t) * M(v) = (x', t'), where M = [m00, m01, m10, m11]
>> is the usual 2x2 matrix-function transform (domain: R,
>> range: 2x2 matrices over R) with the following characteristics.
>
>> [1] (vt,t) * M(v) = (0,t') for some t'. (This is required because we're
>> specifying the second coordinate system moving with velocity v
>> relative to the first.)
>
> Well, t' is not really a requirement in the context of the thread's
> immediate material, where I point out Uncle assAl lies a lot.
Uncle Al needs to be taken at times with a grain of salt, as
do all posters. I for one do not care for his more
racially charged comments, for example. However, those
do not generally enter into the realm of physics, and in
any event there's a lot of history over in that area
of the world -- and lots of bloodletting.
Perhaps he's right. But that's best argued elsewhither.
>
> But, maybe no harm done. We'll see.
>
>> [2] (0,t) * M(v) = (-vt',t') for some t'.
>
> That is just flat, error in the context you presented.
>
> Correct is
>
> [2] (0,t) * (M(v) = (-vt, t') for some t'.
>
> If t'<>t the one thing for sure is that vt<>vt' and -vt is the transform of
> 0.
(0,t) * M(v) = (-vt', t') for some t', as I've said. To stipulate
otherwise invites much stupidity, although one could contemplate
the special case (0,t) * M(v) = (-vt, t), which is simply t = t'
in a slightly different form.
What is velocity? From one standpoint, I can simply
stipulate a fixed point on the moving coordinate system (if
such makes any sense, but usually the origin is attached to
something physical in any real-world measurement -- say,
the prow of a spacecraft) and a pair of points that are
a known distance (say, l units) apart -- known to *me*,
that is; the operator of the spacecraft might wonder
what those two marks mean. I know lightspeed -- or at
least that light has speed -- and can compensate therefor.
Of course this assumes local isotropy, but from MMX local
isotropy is fine; there's no real reason to think that
light will, relative to *me*, travel different speeds when
fired off in different directions.
So I measure. A point on the moving coordinate system will
pass by the origin at time 0, and pass by the other point,
as observed by me, at time lv+l/c. These can be specified
as (0,0) and (0, l/v+l/c), but are more usually specified
as (0,0) and (l, l/v).
In A-space these can be specified (0,0) and (0,t') for some t'.
The observer in O-space don't really care what A-space's t' is.
It turns out this works:
(l, l/v) * [g -vg/c^2 -vg g] = (lg-lg, -lvg/c^2+lg/v) = (0,t')
If one wants to get really picky I can set up a mirror at length
l and do a TWLS. This has some problems -- the lightbeam will
tend to push the mirror away if one contemplates a frictionless
surface (p = E/c, as it turns out, for a photon of energy E).
And of course the laser will take off like a (very weak) rocket.
So usually the laser is bolted to the mirror, which means one has
to compensate for Young's Modulus.
And if the owner of the spacecraft cares to he can mark
off two points on his spacecraft and do similar velocity
measurements. They must equal my own (except for sign
and observational error -- which isn't an issue for such
a thought experiment).
>
>
>> [3] (ct,t) * M(v) = (ct',t') for some t'. (OWLS invariance.)
>
> That, too, is just flat error in the context you presented.
>
> Correct is
>
> [3] (ct,t) * M(v) = (ct-vt,t') for some t'.
In case it has escaped your attention, the LHS is in O-space, but
the RHS is specified in A-space. In any event, with the
Lorentz [g -vg/c^2 -vg g] it turns out that
(ct,t) * [g -vg/c^2 -vg g] = (cgt-vgt, -vg/c+gt) = ((c-v)gt, t').
So in a way (c-v) is still in there. Of course, it turns out
t' = (c-v)gt/c anyway, so x' = ct'.
Remember that this is a mathematical assumption, not a stipulation
of reality. Of course, it turns out SR and GR map darned well to
reality anyway.
>
> ct gives us an x-value which transforms as ct-vt, and with t<>t ct<>ct'.
>
> We need to start over. Way too much error for the rest to be meaningful.
Do you even *know* how to measure? If c = x/t, then c'
= x'/t' (assuming constant lightspeed, anyway); one can't
do c' = x'/t or x/t' as that would be ridiculous.
Note that the logic showing the incompatibility of c' = c and
t' = t works for the pair of statements c' = c and x' = x as well.
In fact, if one stipulates measurement units such that c = 1
lightsecond/second, say (I'm not sure it will ever get adopted,
but one possibility is to use a "nil" -- nano-lightsecond --
as a basis; 1 nil would be 29.9792458 cm, or just under a foot),
then the Lorentz becomes [g -vg -vg g], which is a rather pretty form.
Of course in this new form g = 1/sqrt(1-v^2).
>
> Thanks for the go at it!
You're welcome, but you're going to have to pick your way through
the coordinate transforms *very* carefully.
[.sigsnip]
Dirk Van de moortel
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:621tf2-...@sirius.athghost7038suus.net...
[snip]
> > We need to start over. Way too much error for the rest to be meaningful.
>
> Do you even *know* how to measure? If c = x/t, then c'
> = x'/t' (assuming constant lightspeed, anyway); one can't
> do c' = x'/t or x/t' as that would be ridiculous.
Ghost, the only thing he can do, is a bit of algebra.
He doesn't even know what x and t are supposed
to stand for. He has no idea what the equations
are supposed to mean:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
He talks about "moving values":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
He asks you to "Apply the Lorentz transformation
to the Galilean transfomation equation":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ShowFormula.html
The only thing he wants you to do, is to waste your
time on him ;-)
Dirk Vdm
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
on Mon, 07 Mar 2005 12:29:42 GMT
<WOXWd.1109$2e1...@news.cpqcorp.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:621tf2-...@sirius.athghost7038suus.net...
>> In sci.physics, eleaticus
>> <elea...@bellsouth.net>
>> wrote
>> on Sun, 6 Mar 2005 20:44:14 -0600
>
> [snip]
>
>> > We need to start over. Way too much error for the rest to be meaningful.
>>
>> Do you even *know* how to measure? If c = x/t, then c'
>> = x'/t' (assuming constant lightspeed, anyway); one can't
>> do c' = x'/t or x/t' as that would be ridiculous.
>
> Ghost, the only thing he can do, is a bit of algebra.
> He doesn't even know what x and t are supposed
> to stand for. He has no idea what the equations
> are supposed to mean:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
An interesting notion, that. :-) I wonder what statute.
> He talks about "moving values":
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
Sigh. In SR, v is *supposed* to be constant.
> He asks you to "Apply the Lorentz transformation
> to the Galilean transfomation equation":
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ShowFormula.html
This gets complicated. It's probably similar to Androcle's complaint
that the rod lengthens (because g > 1). However, from an algebraic
standpoint, I've shown that the rod, because it becomes a sheared
band in (x,t) space, the rod *shrinks*, but also *twists* in time.
I suspect even weirder things happen with this area, as it becomes
a tilted post -- and it depends on orientation.
>
> The only thing he wants you to do, is to waste your
> time on him ;-)
As long as it's not money... :-)
>
> Dirk Vdm
eleaticus
> > Ghost, the only thing he can do, is a bit of algebra.
> > He doesn't even know what x and t are supposed
> > to stand for. He has no idea what the equations
> > are supposed to mean:
> > http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
> An interesting notion, that. :-) I wonder what statute.
rofflmfao!
Ghost, have you gone completely nuts? The crime is multiple contradictions
of yourself:
-----------------------------------
Einstein initiates his process by asserting four basic equations and
adds a fifth in his process:
x = ct
x = -ct
x = vt
x' = ct'
x' = -ct'
There are several crimes against logic and science in the assertion of
that set of equations.
-------------------------------
Any ONE of the equations would be OK but any two of the set of three
consititute an absurdity and the second set constitutes an absurdity.
Several SR-cult cretins admitted Einstein was in error on this, probably
including the psychomoron, Dirk, being corrupt enough to imagine that their
godlet didn't do such a thing. I thibnk Dirk was one of them.
Tell us again how you wonder what the crime is, Ghost.
BTW, it being obvious from the word go the Dirk is the slimiest of the
SR-cult cretins on rgp/etc I not once before this ever looked at his crap.
Little did I know he is one of the least talented a junior high students in
the country!
> > He talks about "moving values":
> >
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
> Sigh. In SR, v is *supposed* to be constant.
rofflmfao!
Which part of this material quoted do you think contradicts that, Ghost:
-----------------
Nominally, the velocity v is taken by SR-cultists as being a constant
although in fact what is a positive velocity in one system is a
negative velocity in the other.
------------------------------------------------------------
Do you or do you not know that with the positive directions of the x and x'
axes in the same direction, what the x systems sees as +v for the other
system, is seen by the x' system as -v for the first system?
Hunh? Do you or do you not understand that?
Your response to Dirk's idiocy is moronic.
> > He asks you to "Apply the Lorentz transformation
> > to the Galilean transfomation equation":
> >
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ShowFormula.html
> This gets complicated.
ROFFLMFAO!
No complication at all. The idiot Dirk sees a request - stated two ways -
that 'you' show two DIFFERENT transformations, and the apparent volunteer
idiot, Ghost, makes idiotic agreement with him:
---------------------
Yet, I am the one who can accomplish the task both ways - correctly and the
SR-cult cretin way - and you are the one that would have to violate your
cult's dogma to do it right.
Let A be the general formula in x for the area of a rectangle with its lower
left corner at the origin of x and with its height being h.
Show the formula.
Then transform the formula for A to A', given the simple Newtonian/Galilean
transform x'=x-vt, and using the SR-cult approach to Newtonian/Galilean
transformation of equations .
-----------------------------------------------
So, Ghost, tell us how I was asking for the use of Lorentz in the Galilean
transforms.
For that matter, meet the challenge you cannot meet in the last section.
Innyhoo, I wish I had been following up on Dirk's pandora crap. It is a
marvellous source of SR-cretin idiocy and psychosis material!
The fun thing is any honest person who looks at the stuff realizes he is
psycho.
Why weren't you able to give the material an honest read, Ghost?
eleaticus
David Cross
> This gets complicated. It's probably similar to Androcle's complaint
> that the rod lengthens (because g > 1). However, from an algebraic
> standpoint, I've shown that the rod, because it becomes a sheared
> band in (x,t) space, the rod *shrinks*, but also *twists* in time.
> I suspect even weirder things happen with this area, as it becomes
> a tilted post -- and it depends on orientation.
And to think I thought it was a bit bizarre that in the Lorentz
Transformations you could end up with an angular dependence on the velocity as
you approached the speed of light (think the "rod approaching the hole cut in
the metal" paradox). :-P
--
David Cross
dcross1 AT shaw DOT ca
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:Eu_Wd.7306$Q83....@bignews5.bellsouth.net...
Take a piece a piece of paper.
Draw an x-axis and a t-axis and draw the lines that are
followed by light signals going to your right and to your left.
Write the equations of motion:
The flash to the right has equation
x = c t .
The flash to the left has equation
x = - c t
That's any to of the equations.
According to you that is an absurdity. Congratulations.
Write the equation of motion of a car standing still on your
right at 2 meters:
x = 2
Likewise for another car on your left:
x = -2
According to you that is another absurdity.
Congratulations, you just proved that analytical geometry
is an absurdity.
You are a genius.
Snipped further demonstration of your self-destructive incompetence
and pointer added to entire piece of your meta-crap on your entry
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
Keep'm coming :-)
Dirk Vdm
David Cross
news:Eu_Wd.7306$Q83....@bignews5.bellsouth.net...
> -----------------
> Nominally, the velocity v is taken by SR-cultists as being a constant
> although in fact what is a positive velocity in one system is a
> negative velocity in the other.
> ------------------------------------------------------------
>
> Do you or do you not know that with the positive directions of the x and x'
> axes in the same direction, what the x systems sees as +v for the other
> system, is seen by the x' system as -v for the first system?
Your attitude and language in these posts is most deplorable. However, may I
point out that changing from a positive to a negative is not really a problem
in SR because in the gamma factor the velocity is squared in any case? The
velocity is still going to be frame dependent just as in the Newtonian world.
It's just that the speed of light being an upper limit imposes an additional
restriction on how you must handle relativistic kinematics. As long as one
does not throw in accelerations doing SR does not seem to be a big problem.
:-)
Although I can't resist tweaking you a little bit and asking if you
even -believe- in the Lorentz Transformations.
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:Eu_Wd.7306$Q83....@bignews5.bellsouth.net...
>
Take a piece a piece of paper.
Draw an x-axis and a t-axis and draw the lines that are
followed by light signals going to your right and to your left.
Write the equations of motion:
The flash to the right has equation
x = c t .
The flash to the left has equation
x = - c t
That's any two of the equations.
eleaticus
"David Cross" <nos...@spammenot.com> wrote in message
news:Wp%Wd.594526$Xk.576606@pd7tw3no...
> "eleaticus" <elea...@bellsouth.net> wrote in message
> news:Eu_Wd.7306$Q83....@bignews5.bellsouth.net...
>
> > -----------------
> > Nominally, the velocity v is taken by SR-cultists as being a constant
> > although in fact what is a positive velocity in one system is a
> > negative velocity in the other.
> > ------------------------------------------------------------
> >
> > Do you or do you not know that with the positive directions of the x and
x'
> > axes in the same direction, what the x systems sees as +v for the other
> > system, is seen by the x' system as -v for the first system?
>
> Your attitude and language in these posts is most deplorable.
Did you or did you not notice that first Dirk and then Ghost were saying
stupid, deplorable things?
> However, may I
> point out that changing from a positive to a negative is not really a
problem
> in SR because in the gamma factor the velocity is squared in any case?
The problem/issue in question was the imposition of contra-theoretical
transforms, in the Newtonian case the imposition of t'=t which serves no
purpose but to introduce an extra variable and screwup differential
analyses.
Hence, my mention previously - as quoted by Dirk - of the fact that although
v'<>v you don't impose v'=-v in analyses of SR/Lorentz. There is no
theoretic reason to add t' to the mix but v'=-v is a fact of all setups but
you won't use it. To defend t'=t you must defend not using v'=-v. v'=-v is
not contra-theoretical to any theory; it is just the obvious from the setup
of the systems of axes.
Can you do so?
Further, the full transforms are x'=g(x-vt), where g is either gamma or one
(Newton) and a difference in v is a necessity beause of the vt term, which
is why SRians so often show the reverse-engineered forms, rather than the
actual second system form. In the case of such second system as the
stationary system but still primed we'd have x=g(x'-vt') and it wouldn't
work right unless you reversed the sign of v, which would mean you'd have to
include v' in your differentials approach. (Speaking of the Uncle assAl trsh
he has posted hundreds of times.)
>The
> velocity is still going to be frame dependent just as in the Newtonian
world.
> It's just that the speed of light being an upper limit imposes an
additional
> restriction on how you must handle relativistic kinematics. As long as one
> does not throw in accelerations doing SR does not seem to be a big
problem.
> :-)
That doesn't answer the point about v'=-v. How can you logically exclude an
actual transform and insist on the t'=t transform which only screwsup
correct analysis?
> Although I can't resist tweaking you a little bit and asking if you
> even -believe- in the Lorentz Transformations.
If you believe in them try transforming some B.x = f/x, a typical, general
form.
In its transform according to your wonderful Lorentz transforms, one term we
get is gf/x, but x'=gx implies f/gx. From gf/x we get f/(x/g) which implies
x'=x/g, a length lengthening.
So, if Einstein and Lorentz did such a good job in creating the transform
equation for x to x', why not use it?
Could it be that doing so would horribly screwup the whole mess, so badly
even you would recognize the problem?
Now that you have written 'but that's not how you do it!" try again, this
time responding to the question: if x'=g(x-vt) is such a great idea, why not
use it?
I notice further that you have avoided comment on Dirk's and Ghost's
silliness and callling a request for two transforms a single combination in
one transform.
Why not show a little unbiased candor? Evasion is not honest.
eleaticus
eleaticus
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:MS%Wd.1131$sp1....@news.cpqcorp.net...
Dirk, you cretin. A light packet goes either this way or that way, not both
ways.
It goes at either c or v<>c but not both.
And one does not use x=ct and x=-ct at the same time unless your purpose is
to find out that the equation pair can only apply to c=0 or t=0 or both,
either being a strange limitation for transformations for any time value,
and a c > 0.
Similarly, you do not use x=ct and x=vt simultaneously unless it is to find
out that v=c.
Even now - after a decade - it is a matter of increasing wonder that people
nominally intelligent enough to get past the third grade cannot - flat will
not - see the obvious.
eleaticus
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:fN2Xd.7723$Q83....@bignews5.bellsouth.net...
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
>
> It goes at either c or v<>c but not both.
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
>
> And one does not use x=ct and x=-ct at the same time
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
> unless your purpose is
> to find out that the equation pair can only apply to c=0 or t=0 or both,
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
> either being a strange limitation for transformations for any time value,
> and a c > 0.
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
>
> Similarly, you do not use x=ct and x=vt simultaneously unless it is to find
> out that v=c.
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
>
> Even now - after a decade - it is a matter of increasing wonder that people
> nominally intelligent enough to get past the third grade cannot - flat will
> not - see the obvious.
Brilliant.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
http://users.pandora.be/vdmoortel/dirk/Stuff/WebsterAtSchool.gif
Listen punkie, I have explained all this to Marcel Luttgens several
years ago several hundreds of times. You should have paid attention.
Use google or email him - he will explain. Wish him all the best and
a lot of courage.
Don't come back before you apologized to analytical geometry.
Dirk Vdm
The Ghost In The Machine
<Eu_Wd.7306$Q83....@bignews5.bellsouth.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:1l5uf2-...@sirius.athghost7038suus.net...
>> In sci.physics, Dirk Van de moortel
>> <dirkvand...@ThankS-NO-SperM.hotmail.com>
>
>> > Ghost, the only thing he can do, is a bit of algebra.
>> > He doesn't even know what x and t are supposed
>> > to stand for. He has no idea what the equations
>> > are supposed to mean:
>> > http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
>
>> An interesting notion, that. :-) I wonder what statute.
>
> rofflmfao!
>
> Ghost, have you gone completely nuts? The crime is multiple
> contradictions of yourself:
Since when is that illegal? As it is, I see no contradiction here.
> -----------------------------------
> Einstein initiates his process by asserting four basic equations and
> adds a fifth in his process:
>
> x = ct
> x = -ct
> x = vt
>
> x' = ct'
> x' = -ct'
OK, we have x, t, x', t', as variables, and c and v as constants.
Of course your system is inconsistent anyway (x = -ct and x = +ct
cannot both be true unless either c=0 or t=0, which is relatively
uninteresting), so let's be a little more logical and assert
the following, where I've bastardized Irving M. Copi's notation
to fit ASCII's limitations. (I've omitted the qualifiers that
everything is a real number for brevity.)
(Av)(EA)(EB)(EC)(ED)(
[1] (Ax)(At)(x' = Ax + Bt)
[2] (Ax)(At)(t' = Cx + Dt)
[3] (Ax')(At')(x = Ax' - Bt')
[4] (Ax')(At')(t = Cx' - Dt')
[5] (At)(Act+Bt = c(Cct + Dt))
[6] (At)(Avt + Bt = 0)
)
The first two are a consequence of space isotropy, or an assumption
that the transform must be linear. The next two are the assumption
that, if observer X is moving from observer Y with velocity +v,
then observer Y must be moving from observer X with velocity -v.
#5 is from the constancy of OWLS. #6 is the observation that the
origin of the second observer must be moving from the first observer
with velocity v.
I'm not sure I've specified this quite right but it's pretty close.
>
> There are several crimes against logic and science in the assertion of
> that set of equations.
> -------------------------------
>
> Any ONE of the equations would be OK but any two of the set of three
> consititute an absurdity and the second set constitutes an absurdity.
> Several SR-cult cretins admitted Einstein was in error on this, probably
> including the psychomoron, Dirk, being corrupt enough to imagine that their
> godlet didn't do such a thing. I thibnk Dirk was one of them.
>
> Tell us again how you wonder what the crime is, Ghost.
>
> BTW, it being obvious from the word go the Dirk is the slimiest of the
> SR-cult cretins on rgp/etc I not once before this ever looked at his crap.
> Little did I know he is one of the least talented a junior high students in
> the country!
>
>> > He talks about "moving values":
>> >
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
>
>> Sigh. In SR, v is *supposed* to be constant.
>
> rofflmfao!
>
> Which part of this material quoted do you think contradicts that, Ghost:
> -----------------
> Nominally, the velocity v is taken by SR-cultists as being a constant
> although in fact what is a positive velocity in one system is a
> negative velocity in the other.
Not sure what you mean by "system" here. Did you mean "observer"?
In which case, you are entirely correct -- if B is moving with
velocity +v from A, then A is moving with velocity -v from B.
But v is still a constant *if one sticks with one observer*.
You weren't. Is this an issue?
>
> For that matter, meet the challenge you cannot meet in the last section.
If the area at rest of the rectangle is A, and the rectangle is
oriented parallel to the x acis, then the the area of the rectangle
in B-space is going to be A/g -- if it makes sense to make any
sort of statement regarding the rectangle's area at all.
Since y = y' we can ignore the vertical part of the rectangle
and concentrate on the rod.
Now, what is the length of the rod along the x axis? In A-space,
one envisions two points (0,t) and (x,t), where x is the length
of the rod. In B-space, these transform into (-vtg, tg)
and (x-vtg,(t-vx/c^2)g). However, note that the rod has twisted
in (x,t). Is this an accurate measurement?
We can instead envision two points on the rod in A-space
(0,t) and (x,T), where we'll figure out T later.
These transform into (-vtg,tg) and (x-vTg, (T-vx/c^2)g).
If B equates (T-vx/c^2)g = tg, then T = t+vx/c^2, and
we get the two points (-vtg, tg) and (x-v(t+vx/c^2)g, tg), which
means B will measure (x-vt-xv^2/c^2)g+vtg = x(1-v^2/c^2)g = x/g
as the length of the rod (remember that g = 1/sqrt(1-v^2/c^2)
so 1-v^2/c^2 = 1/g^2).
Confusing? You betcha! The best one can do in this space is
that B will take a certain amount of time to observe the
rod's endpoints move past him -- namely, x/(vg) seconds.
This is not a length, but a time.
If one asserts x' = 0 = (x-vt)g, then t = x/v. This is
in A-space, of course, and A cannot really observe it
properly (as the light will take time to propagate from the
far end of the rod), although he might try. In B-space,
t' = (t-vx/c^2)g = (x/v-vx/c^2)g = (x/v)(1-v^2/c^2)g = x/(vg).
But has the rod really contracted? That's admittedly a
question for philosophical debate, despite B's measurements
on the moving rod.
>
> Innyhoo, I wish I had been following up on Dirk's pandora crap. It is a
> marvellous source of SR-cretin idiocy and psychosis material!
>
> The fun thing is any honest person who looks at the stuff realizes he is
> psycho.
>
> Why weren't you able to give the material an honest read, Ghost?
And which material are you referring to?
>
>
> eleaticus
>
>
>> #191, ewi...@earthlink.net
>> It's still legal to go .sigless.
>
>
--
The Ghost In The Machine
<fN2Xd.7723$Q83....@bignews5.bellsouth.net>:
Strawman argument. Let's see if this helps.
Assume two coordinate systems, (x,t) and (X,T), and two constants
c and v, 0 < v < c [*], and that
[1] For any x, X, t, and T, X = Ax + Bt .
[2] For any x, X, t, and T, T = Cx + Dt .
We further stipulate that:
[3] For any x, X, t, and T, if x = vt then X = 0.
[4] For any x, X, t, and T, if X = -vT then x = 0.
[5] For any x, X, t, and T, if x = ct then X = cT.
and for consistency's sake, we can also assume that
[6] For any x, X, t, and T, if X = Ax + Bt and T = Cx + Dt,
then x = AX - BT and t = -CX + DT.
One can then prove that A = g, B = -vg, C = -vg/c^2, and D = g,
if I've specified this correctly.
One can of course quibble as to the postulates; one issue for
instance is the lightspeed invariance. However, as far as I
know most lightspeed measurements have not measured c * 0.9999
or c * 1.0001 as one might expect -- the Earth moves around Sol
at about 0.0001 * c -- or perhaps c * 0.9989 or c * 1.0011 as
Sol moves around the Galaxy at about 0.001 * c, but plain
old-fashioned ordinary c.
Every time, with the possible exception of Galileo's shutter lanterns.
:-)
(There's one that's a bit off -- again by Galileo -- but most
historians suggest that the error is because Galileo had the
wrong distance to Jupiter.)
[*] this isn't all that necessary but imaginary coordinates look a bit
weird in a physical context.
--
eleaticus
You first deny I had a point then completely verify my point. Are you
actually a 'multiple personality' psycho?
eleaticus
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
message news:k7kvf2-...@sirius.athghost7038suus.net...
eleaticus
> In sci.physics, eleaticus
> <elea...@bellsouth.net>
> >> > -----------------------------------
> >> > Einstein initiates his process by asserting four basic equations and
> >> > adds a fifth in his process:
> >> > x = ct
> >> > x = -ct
> >> > x = vt
> >> > x' = ct'
> >> > x' = -ct'
> >> > There are several crimes against logic and science in the assertion
of
> >> > that set of equations.
> >> > -------------------------------
> >> > Any ONE of the equations would be OK but any two of the set of three
> >> > consititute an absurdity and the second set constitutes an absurdity.
> > Dirk, you cretin. A light packet goes either this way or that way, not
both
> > ways.
> > It goes at either c or v<>c but not both.
> > And one does not use x=ct and x=-ct at the same time unless your purpose
is
> > to find out that the equation pair can only apply to c=0 or t=0 or both,
> > either being a strange limitation for transformations for any time
value,
> > and a c > 0.
> > Similarly, you do not use x=ct and x=vt simultaneously unless it is to
find
> > out that v=c.
> Strawman argument. Let's see if this helps.
Ghost, why try so hard to join the SR-cretin Hall of Fame?
> Assume two coordinate systems, (x,t) and (X,T), and two constants
> c and v, 0 < v < c [*], and that
> [1] For any x, X, t, and T, X = Ax + Bt .
> [2] For any x, X, t, and T, T = Cx + Dt .
Explain what that has to do with Einstein's use simultaneously, in one
continuous derivation, of:
x' = ct
x' = -ct
x' = vt
BTW, what happened to your complete nonsense in this:
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
message news:s5jsf2-...@sirius.athghost7038suus.net...
> If one assumes, mathematically, that c' = c, then one has to
> at least consider the possibility that t' != t.
OK assumption for the heck of it, but MMX fits c'=c+v.
> Assume that (x,t) * M(v) = (x', t'), where M = [m00, m01, m10, m11]
> is the usual 2x2 matrix-function transform (domain: R,
> range: 2x2 matrices over R) with the following characteristics.
> [1] (vt,t) * M(v) = (0,t') for some t'. (This is required because we're
> specifying the second coordinate system moving with velocity v
> relative to the first.)
Well, t' is not really a requirement in the context of the thread's
immediate material, where I point out Uncle assAl lies a lot.
But, maybe no harm done. We'll see.
> [2] (0,t) * M(v) = (-vt',t') for some t'.
That is just flat error in the context you presented.
Correct is
[2] (0,t) * (M(v) = (-vt, t') for some t'.
If t'<>t the one thing for sure is that vt<>vt' and -vt is the transform of
0.
> [3] (ct,t) * M(v) = (ct',t') for some t'. (OWLS invariance.)
That, too, is just flat error in the context you presented.
Correct is
[3] (ct,t) * M(v) = (ct-vt,t') for some t'.
ct gives us an x-value which transforms as ct-vt, and with t'<>t ct<>ct'.
We need to start over. Way too much error for the rest to be meaningful.
Thanks for the go at it!
eleaticus
The Ghost In The Machine
<AekXd.27355$%Y4....@bignews6.bellsouth.net>:
What, precisely, is your objection to the Lorentz?
>
>> Assume two coordinate systems, (x,t) and (X,T), and two constants
>> c and v, 0 < v < c [*], and that
>
>> [1] For any x, X, t, and T, X = Ax + Bt .
>> [2] For any x, X, t, and T, T = Cx + Dt .
>
> Explain what that has to do with Einstein's use simultaneously, in one
> continuous derivation, of:
>
> x' = ct
> x' = -ct
> x' = vt
Would you prefer different variable names and a functor/tensor operator
of some sort?
I could stipulate, for example, something along the following lines:
Given the set S: {(cu,u): u in R}, S * L(v) = S
Given the set T: {(vt,t): t in R}, S * L(v) = {(0,t): t in R}
where L(v) is the Lorentz matrix (expressed as a function of v)
and S * L(v) is the set-extension of (x,y) * L(v);
I'm deriving, mathematically speaking, the operation:
*: domain: (set of R x R) x (matrix over R)
range: set of R x R
from the operation:
*: domain: (R x R) x (matrix over R)
range: R x R
>
> BTW, what happened to your complete nonsense in this:
Nothing. It's been archived for all to see. I don't put
X-No-Archive in my posts.
(Of course, others might quibble as to which is the
more nonsensical: my post or your replies thereto.)
>
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:s5jsf2-...@sirius.athghost7038suus.net...
>> If one assumes, mathematically, that c' = c, then one has to
>> at least consider the possibility that t' != t.
>
> OK assumption for the heck of it, but MMX fits c'=c+v.
That it does. There is no conceivable method by which MMX can
ever hope to differentiate between the two conditions:
[1] ballistic theory (where light behaves as little hockey pucks
on a frictionless "surface" of some sort, or as little bullets
flying in a Newtonian vacuum)
[2] SR
In order to differentiate between these two one must perform
other experiments -- the simplest might be to use the MMX but
point its intake at, say, Venus, which is moving with respect
to the Earth. Many other experiments are also possible, and
have already been done.
>
>> Assume that (x,t) * M(v) = (x', t'), where M = [m00, m01, m10, m11]
>> is the usual 2x2 matrix-function transform (domain: R,
>> range: 2x2 matrices over R) with the following characteristics.
>
>> [1] (vt,t) * M(v) = (0,t') for some t'. (This is required because we're
>> specifying the second coordinate system moving with velocity v
>> relative to the first.)
>
> Well, t' is not really a requirement in the context of the thread's
> immediate material, where I point out Uncle assAl lies a lot.
>
> But, maybe no harm done. We'll see.
>
>> [2] (0,t) * M(v) = (-vt',t') for some t'.
>
> That is just flat error in the context you presented.
Why?
>
> Correct is
>
> [2] (0,t) * (M(v) = (-vt, t') for some t'.
>
> If t'<>t the one thing for sure is that vt<>vt' and -vt is
> the transform of 0.
Well, I for one might dispute that but I frankly don't see
the problem here -- or, for that matter, how one expects
to do a measurement in A-space with a time interval in B-space.
>
>
>> [3] (ct,t) * M(v) = (ct',t') for some t'. (OWLS invariance.)
>
> That, too, is just flat error in the context you presented.
>
> Correct is
>
> [3] (ct,t) * M(v) = (ct-vt,t') for some t'.
I hope that you're seeing that I'm *postulating* lightspeed
invariance. In order to postulate such one must be able to
measure the speed in both measurement-systems. You're
mixing lengths in A-space and time in B-space again.
>
> ct gives us an x-value which transforms as ct-vt, and with t'<>t ct<>ct'.
>
> We need to start over. Way too much error for the rest to be meaningful.
>
> Thanks for the go at it!
>
> eleaticus
>
>
The Ghost In The Machine
<b9kXd.27351$%Y4.2...@bignews6.bellsouth.net>:
> Ghost, what was your purpose in this post?
Apparently to annoy you.
>
> You first deny I had a point then completely verify my point. Are you
> actually a 'multiple personality' psycho?
You had a point? Well, never mind; let's try a different approach.
Again, assume two observers A and B. Observer A has a light source
and a camera. Observer B has a light source and camera.
Observer A has a ruler of a known length. Observer B has a ruler
of the same length (if both are compared in a common reference-frame).
Observer A has a clock. Observer B has a clock of identical make.
If we assume that Observer A and Observer B are physically coincident
at time 0 and that Observer B is moving uniformly with respect
to observer A with velocity v (the simplest method I can think of
verifying this is for Observer A to place his ruler along B's path
and measure how long it takes for Observer B's origin to pass
by each endpoint, compensating for the light-speed time it takes
for lightpulses from B to come back to A's origin), that light
is always moving at constant speed (if properly measured), and that
neither clock is affected by the acceleration necessary to set
up this thought-experiment what can we conclude?
If B's ruler is plotted in A's (x_a, t_a) coordinate system, one
gets a *band*, which looks a bit like a leaning parallelogram.
A's ruler will look like a rectangle, of course.
A's ruler of course can be described as R_AA = {(x,t): 0 <= x <= l}.
A can also fire a lightbeam. This beam, neglecting such issues as
Airy discs and photon packet widths, can be modeled using
the equation Bfwd_AA = {(x,t): x = ct, x in R, t in R}, where c is
our assumed universal lightspeed constant. A slightly shorter
variant is Bfwd_AA = {(ct,t): t in R}. One can also fire a lightbeam
in the other direction: Brev_AA = {(-ct,t): t in R}. The
forward direction, however, should be sufficient.
It is now time for our tensor. For the purposes of this post [*], a
tensor is an entity that is a 1-1 mapping from one vector space
to another vector space (in our case, both spaces are 2-dimensional
over R). I will use '*' here; the more conventional notation looks
a bit like a circled x.
(x,t) * L(v) = (x', t')
where (x,t) is one coordinate point, and (x', t') is some other
coordinate point, and L(v) is a function mapping a real to
a tensor. More on what precisely L(v) is, later.
Some examples of tensors:
One tensor simply shrinks everything in half:
(x,t) * Half = (x/2,t/2)
Another will do a shear transformation -- or, if one prefers,
a Galilean velocity move:
(x,t) * Shear(v) = (x - vt, t)
Another might swap time and space -- or at least the two
coordinates we interpret as such.
(x,t) * Swap = (t,x)
Still another might do something extremely silly (but still 1-1):
(x,t) * Weird = (x^3, cuberoot(t))
Some of these might have physical meaning; most of them will not.
At this point it's pure math.
We can extend the operator '*' in the fairly obvious fashion:
S * L(v) = {s * L(v): s in S}
where S is an arbitrary set of real tuples (i.e., a subset of R^2),
and we can now ask questions such as how various subsets of R^2
look after the tensor has munged them. For example, the
unit square, which we might represent as [0,1] x [0,1],
will change into a parallelogram after Shear(v) is
through with it; one might describe the result thusly:
([0,1] x [0,1]) * Shear(-2) = the solid polygon/parallelogram with
endpoints (0,0), (1,0), (3,1), and (2,0).
It turns out that
([0,1] x [0,1]) * Swap = ([0,1] x [0,1]) * Weird = [0,1] x [0,1].
And of course ([0,1] x [0,1]) * Half = [0,1/2] x [0,1/2].
For purposes of this post a ruler at rest can be thought
of as the set [0,l] x R, for some l. (This is, of course,
a thought-ruler; we don't go into detail on how one
cuts the ruler from the tree, or creates the markings, or
whether it will be destroyed in a fire sometime in the 23rd
century, gets broken in half by a careless student, etc. etc.)
We assume the following for purposes of this argument.
[1] {(vt,t): t in R} * L(v) = {(0,t): t in R}
(This is basically a requirement that B is moving with velocity v
with respect to A, mathematically speaking.)
[2] {(ct,t): t in R} * L(v) = {(ct,t): t in R}
(Read that line *very* carefully! It means that the two *sets*
are equal, not that the point (ct,t) gets mapped to the point
(ct,t).)
[3] (a,b) * L(v) + (c,d) * L(v) = ( (a,b) + (c,d) ) * L(v)
where '+' is the usual vector addition. (There are more
thorough treatments which can discard this step, but I for
one would have to dig. The assumption here is that the
space in this thought-experiment is isotropic: the same
no matter where one is positioned.)
[4] (0,0) * L(v) = (0,0)
This is primarily for convenience -- and actually isn't even strictly
necessary, as one can deduce it from [3]:
((a,b) - (a,b)) * L(v) = (0,0) * L(v)
((a,b) - (a,b)) * L(v) = (a,b) * L(v) - (a,b) * L(v) = (0,0) QED
We also have to define yet another operator. If we have a point
P in R^2 and two tensors J and K, then we want to define '*' such
that
(P * J) * K = P * (J * K)
where J * K is *another* tensor. Admittedly, I might do better
here with another character (e.g., a small circle), but for
now this shouldn't be too confusing.
We also define the identity tensor (x,t) * I = (x,t), and make
an assumption that
[5] L(v) * L(-v) = I
and finally, that
[6] L(0) = I
though that's probably not strictly necessary anyway.
So, given all this, what is L(v)?
Another notation might be handy; since L(v) is known linear
it has to be a matrix. I stipulate here without proof that
the tensor
L(v) = [g -vg -vg/c^2 g]
where g = 1/sqrt(1-v^2/c^2), fulfills all these requirements.
Note that the more conventional notation
x' = (x - vt)*g
t' = (t - vx/c^2)*g
can be expressed (x,t) * L(v) = (x',t').
Now: what, precisely, is your objection?
[snip for brevity]
[*] I'm not really *that* familiar with tensors, and tensors are
actually overkill for SR-type problems. However, tensors
are routinely used in GR.
(And hopefully I'm doing this stuff at least somewhat right,
from an introductory level.)
eleaticus
> > You first deny I had a point then completely verify my point. Are you
> > actually a 'multiple personality' psycho?
> You had a point? Well, never mind; let's try a different approach.
Yes, asshole. The point was always obvious and flatly stated. Einstein's use
of x=ct, x=-ct, and x=vt where the derivation was supposed to be for all
values of x, any t>=0, and any |v|<|c| was absurd. And how did you respond?
First:
>Since when is that illegal? As it is, I see no contradiction here.
And, btw, I said it was ok to use the contradictories to see if there was a
common solution.
Then:
>Of course your system is inconsistent anyway (x = -ct and x = +ct
>cannot both be true unless either c=0 or t=0, which is relatively
>uninteresting),
ROFFLMFAO!
Einstein's derivation depended on HIS system in which c>0 (and t>0 in most
cases).
Asshole!
eleaticus
eleaticus
message news:cukvf2-...@sirius.athghost7038suus.net...
> >> > -----------------------------------
> >> > Einstein initiates his process by asserting four basic equations and
> >> > adds a fifth in his process:
> >> > x = ct
> >> > x = -ct
> >> > x = vt
> >> > x' = ct'
> >> > x' = -ct'
> >> > There are several crimes against logic and science in the assertion
of
> >> > that set of equations.
> >> > -------------------------------
> >> > Any ONE of the equations would be OK but any two of the set of three
> >> > consititute an absurdity and the second set constitutes an absurdity.
> Assume two coordinate systems, (x,t) and (X,T), and two constants
> c and v, 0 < v < c [*], and that
> [1] For any x, X, t, and T, X = Ax + Bt .
> [2] For any x, X, t, and T, T = Cx + Dt .
Explain what that has to do with Einstein's use simultaneously, in one
continuous derivation, of:
x = ct
x = -ct
x = vt
('typos' corrected, the originals being quoted many times above).
Since you so readily endorsed Dirk The Dim's idiocy about those three
assertions Einstein made in his 1916/1962 derivation of his transforms, why
don't you explain how a valid derivation can be made by simultaneous use of
three mutually contradictory assertions?
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet the only equation (plus dengenerate versions) that is
actually invariant under Einstein's xforms is x^2+y^2+z^2-(ct)^2.
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet x=ct and x=-ct have common solutions only where c=0 or t=0.
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet x=ct and x=vt have common solutions only where v=c or t=0.
Soon after Dirk The Dim's 33rd birthday his Mommy decided to let the lad
take a trip on his own. A family friend owned a hotel in a nearby town and
promised to look after the boy.
Happily, Dimmy arrived at the hotel but soon was crestfallen. He had
forgotten his pencils. Oh, shoot! He had so terribly wanted to figure out
whether or not it was really true that 3+2=5.
He asked the hotel desk clerk if there was any place nearby where he could
buy pencils. At first the clerk was reluctant to answer but he checked with
the owner and was told it was ok to tell the lad where it could be done, but
the clerk would have to chaperon the boy.
"Go out the front doors and turn left, Dimmy, and one block away you can buy
pencils at only $1.20 a dozen at the convenience store."
"Ohh!" exclaimed the lad, "that will kill my allowance!"
"Or, you can turn right and buy pencils for $1.25 a dozen at the drugstore."
Dimmy decided a drugstore's pencils would surely be worth the extra nickle a
dozen so the clerk took him by the hand and helped him get there and back.
Dimmy so so overwhelmed by the adventure that he couldn't concentrate on
what he had thought was a major problem, is 2+3 equal to 5.
What, he wondered, is the mathematical law governing pencil prices in this
town? Surely there must be one.
Let the total cost C be 1.20D in the one direction, he thought, and 1.25D in
the other direction.
That gives us C=1.20D and C=1.25D, he continued, and 2C = 2.45D, and
C=1.225D.
Wow! he thought, I did it! The underlying law of pencil prices in this town!
Mommy will be so proud!
Little did Dimmy understand, later, why people said you can't use
contradictories simultaneously in a continuous derivation. That their only
use simultaneously was to see if there was a common solution. That any
derivation could only apply to the common values, if any.
Some even said that a whole logical proof technique was based on finding
just one pair of contradictories, reduction to the absurd.
But Dimmy knew better. His only regret was that there hadn't been a third
source for pencils, maybe one for C=1.00D, which would have given him
C=1.15D, a much more elegant and beautiful mathematical law!
eleaticus
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:1x0Yd.22440$6g7....@bignews1.bellsouth.net...
Heh... so you have a problem with negative time as well ;-)
When he returns, Androcles will be pleased to finally have
found a worthy ally:
Negative time? Impossible!:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegTime.html
Moving Coordinates & Negative time revisited:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegTime2.html
Stop writing equations, Webster, you're going to hurt yourself.
Dirk Vdm
David Cross
wrote:
>The transforms are supposed to be valid for all values of x, any t>=0, and
>any |v|<|c|, yet the only equation (plus dengenerate versions) that is
>actually invariant under Einstein's xforms is x^2+y^2+z^2-(ct)^2.
No problem with that. The spacetime interval is the invariant quantity, and
you can prove that with a spacetime diagram.
Think of the familiar example of the city whose map differs by a rotation. The
distance between two points doesn't depend on the coordinate system but the
identification of where they are relative to some arbitrary x, y and z axes
does depend on the coordinate system.
No mystery!
---
eleaticus
"David Cross" <spamd...@nospam.net> wrote in message
news:qqa131loikm0fop6g...@4ax.com...
Funny thing about that. I did look at a map just now of 'my town' and
compared x^2+y^2-(ct)^2 at two different times with the measurable distance
the map showed. There was no agreement, no invariance whatsover!
Your illustration is just irrelevant idiocy. Standard idiocy but idiocy
nonetheless.
> No mystery!
You're right. x^2+y^2+z^2 is the square of a distance but subtract (ct)^2
and you get complete idiocy, not even the square of the distance between
(x,y,z) and ct.
eleaticus
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:sh2Yd.20443$5T6....@bignews4.bellsouth.net...
>
> "David Cross" <spamd...@nospam.net> wrote in message
> news:qqa131loikm0fop6g...@4ax.com...
> > On Thu, 10 Mar 2005 12:41:02 -0600, "eleaticus" <elea...@bellsouth.net>
> > wrote:
> >
> > >The transforms are supposed to be valid for all values of x, any t>=0,
> and
> > >any |v|<|c|, yet the only equation (plus dengenerate versions) that is
> > >actually invariant under Einstein's xforms is x^2+y^2+z^2-(ct)^2.
> >
> > No problem with that. The spacetime interval is the invariant quantity,
> and
> > you can prove that with a spacetime diagram.
> >
> > Think of the familiar example of the city whose map differs by a rotation.
> The
> > distance between two points doesn't depend on the coordinate system but
> the
> > identification of where they are relative to some arbitrary x, y and z
> axes
> > does depend on the coordinate system.
>
> Funny thing about that. I did look at a map just now of 'my town' and
> compared x^2+y^2-(ct)^2 at two different times with the measurable distance
> the map showed. There was no agreement, no invariance whatsover!
>
> Your illustration is just irrelevant idiocy. Standard idiocy but idiocy
> nonetheless.
Perhaps, just like Androcles, you haven't got the foggiest
idea about what coordinates actually are.
Perhaps someone should take a hammer and jam a tiny
bit of analytical geometry into your head.
But given the amount of water that must be in there... well,
let's say I'm a bit pessimistic.
Dirk Vdm
David Cross
wrote:
>> No problem with that. The spacetime interval is the invariant quantity,
>and
>> you can prove that with a spacetime diagram.
>>
>> Think of the familiar example of the city whose map differs by a rotation.
>The
>> distance between two points doesn't depend on the coordinate system but
>the
>> identification of where they are relative to some arbitrary x, y and z
>axes
>> does depend on the coordinate system.
>
>Funny thing about that. I did look at a map just now of 'my town' and
>compared x^2+y^2-(ct)^2 at two different times with the measurable distance
>the map showed. There was no agreement, no invariance whatsover!
You seem to just love doing this deliberate misinterpretation BS, don't you?
Hello? I was using an a-n-a-l-o-g-y. What is an analogy? It is:
a·nal·o·gy ( P ) Pronunciation Key (-nl-j)
n. pl. a·nal·o·gies
Similarity in some respects between things that are otherwise dissimilar.
A comparison based on such similarity. See Synonyms at likeness.
The analogy is that the Pythagorean-derived distance is the invariant when
using space coordinates only. In spacetime, the interval is the invariant.
YOU ARE SO FIRED, ELEATICUS. FIRED, YOU HEAR ME?
Tom Capizzi
How was Einstein making any assertions in 1962? through a oiuja board?
> don't you explain how a valid derivation can be made by simultaneous use
> of
> three mutually contradictory assertions?
>
> The transforms are supposed to be valid for all values of x, any t>=0, and
> any |v|<|c|, yet the only equation (plus dengenerate versions) that is
> actually invariant under Einstein's xforms is x^2+y^2+z^2-(ct)^2.
>
> The transforms are supposed to be valid for all values of x, any t>=0, and
> any |v|<|c|, yet x=ct and x=-ct have common solutions only where c=0 or
> t=0.
>
> The transforms are supposed to be valid for all values of x, any t>=0, and
> any |v|<|c|, yet x=ct and x=vt have common solutions only where v=c or
> t=0.
>
[snip irrelevant garbage]
> eleaticus
>
>
>
>
Dirk Van de moortel
"Tom Capizzi" <etian...@verizon.net> wrote in message news:1i3Yd.86394$sR5.52258@trndny05...
Common and well known troll tactic.
They hope that you make a remark (or even better, a joke)
about a detail, so they can start whining about your picking
out a little detail in order to ignore the general idea.
[ This was lesson #13 of my free course on Applied Village
Idiot Psychology (AVIP) ]
Dirk Vdm
Tom Capizzi
as an ineducable bore and will not respond further.
The Ghost In The Machine
<oE0Yd.22443$6g7....@bignews1.bellsouth.net>:
So OK then.
You have now disproved SR. Basically, SR math has now
been invalidated, and any grand theories based thereon
(GR, string/brane, Standard Model, and other such) will
have to be completely reworked.
Please propose your theory. Include mathematical predictions.
Dirk Van de moortel
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:iib7g2-...@sirius.athghost7038suus.net...
Before he does the latter, first Descartes will have to
start reworking Analytic Geometry as well.
Who do we ask to rework Linear Algebra?
Dirk Vdm
The Ghost In The Machine
on Fri, 11 Mar 2005 09:23:11 GMT
<3sdYd.1510$Zd4...@news.cpqcorp.net>:
Androcles, maybe? ;-)
>
> Dirk Vdm
jahn
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:iib7g2-...@sirius.athghost7038suus.net...
Two particle pairs in relative motion violates the first postulate
of SR. The magnets would fall off of your refrigerator if it didn't.
Two coupled structures in the near-field violate the second
postulate of SR. The transistiors in your computer would stop
working if it didn't.
The use of visible light at atomic wavelengths coupled to
structures that change form at visible light frequencies (atoms)
must surely be the most exciting way to study the near and far
field anomalies of Maxwell's equations. ( The reason for SR? )
For my own personal tastes, I rank the study technique
as every bit as exciting as self mutilation, influenza or up
close and personal poison plant study.
With all those charges and waves whizzing around, I hope
I am forgiven for not quite understanding who or what proves
or disproves anything.
Sue...
Mich
Thanks in advance for your answers.
The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in
message news:k7kvf2-...@sirius.athghost7038suus.net...
> In sci.physics, eleaticus
> <elea...@bellsouth.net>
> > -----------------------------------
> > Einstein initiates his process by asserting four basic equations and
> > adds a fifth in his process:
> >
> > x = ct
> > x = -ct
> > x = vt
> >
> > x' = ct'
> > x' = -ct'
>
> OK, we have x, t, x', t', as variables, and c and v as constants.
My first question would be; if c and v are velocity vectors as opposed to
speed,since it can either be + or -, why isn't x a displacement as opposed
to a distance, which ought to mean that if x = -ct, then, -x = ct?
...I think I will not go any farther than this for now.
Andre
The Ghost In The Machine
<mi...@efni.com>
wrote
on Sat, 12 Mar 2005 09:24:22 -0500
<1135und...@corp.supernews.com>:
All values are scalars, though if you like one can move into the
vector realm by replacing x with X or P (don't use 'p'; that's too
easily confused with momentum :-) ). Note, however, that the
Lorentz rotation/contraction/transformation only rotates along
the direction of travel, which is by convention taken to be the
x-axis, and the time coordinate. In ASCII, this is conventionally
rendered
x' = (x-vt) * gamma
y' = y
z' = z
t' = (t-vx/c^2) * gamma
where gamma = 1/sqrt(1-v^2/c^2).
In the case of the light beam, one should be able to use
any velocity whose magnitude is c, though I'm not sure
what will happen from an algebraic standpoint. It might
be mildly interesting, and Einstein does make the implicit
assumption that one can in fact rigidly rotate the coordinate
system to the desired orientation -- an assumption that is,
at best, premature, though not all that problematic for
most physics problems of the "let a train be moving along
the x axis" sort.
As for the x=ct / x=-ct / x=vt controversy, that's probably
best resolved by using subscripted variables or some such.
Einstein, in his thesis (an English transcript is at
http://www.fourmilab.ch/etexts/einstein/specrel/www/ )
apparently used x,y,z,t for coordinates in the stationary
system (termed 'K') and xi (ξ), eta (η),
zeta (ζ), and tau (τ) for coordinates in the
moving system (termed 'k'). He then proceeds to use notations
such as x' = x-vt, tau(x',y,z,t), and partial derivatives,
ultimately arriving at the result
xi = (x-vt) * gamma
eta = y
zeta = z
tau = (t-vx/c^2) * gamma
by proving that phi(t) (ϕ, φ, or ɸ) = 1.
The confusion starts when one replaces the greek xi, eta, zeta, tau
with x', y', z', and t' (a proper replacement of Einstein's x'
might be x_m or some such, were one to redo this analysis
using pure ASCII).
It gets weirder.
The Einstein transcription uses beta (β) for what we today
would use gamma (γ). See
http://scienceworld.wolfram.com/physics/SpecialRelativity.html
for such an example.
All this suggests that one had better be very careful in declaring
what the entities mean beforehand.
>
> Andre
The Ghost In The Machine
<susyse...@yahoo.com.au>
wrote
on Sat, 12 Mar 2005 03:03:50 -0500
<39fm3tF...@individual.net>:
I was being sarcastic. :-)
However, you have now given several examples of a disproof of SR,
or perhaps more properly examples outside of the realm of SR.
Do they fall outside or disprove GR?
And, do they fall outside or disprove Newton's "laws of motion"?
BTW: visible light is not at atomic wavelengths. Visible light
is in the range of maybe 380 nm to 760 nm. The size of an atom is
about 100-200 pm. You're talking very hard X-rays, almost
gamma rays.
http://mywebpages.comcast.net/bondono2/ispect.html
However, the electron binding energy for, say, neon, is 21.6 eV,
which translates into the soft X-ray range -- and usually
one does't even need that much energy, as one need merely
transition a bound electron (in the case of neon one might
use a 2p orbital) to another, higher-energy state, then let
the electron transition back emitting a photon in the process.
[rest snipped]
jahn
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:1kmbg2-...@sirius.athghost7038suus.net...
[snip]
> >>
> >> So OK then.
> >>
> >> You have now disproved SR. Basically, SR math has now
> >> been invalidated, and any grand theories based thereon
> >> (GR, string/brane, Standard Model, and other such) will
> >> have to be completely reworked.
> >
> > Two particle pairs in relative motion violates the first postulate
> > of SR. The magnets would fall off of your refrigerator if it didn't.
> >
> > Two coupled structures in the near-field violate the second
> > postulate of SR. The transistiors in your computer would stop
> > working if it didn't.
> >
> > The use of visible light at atomic wavelengths coupled to
> > structures that change form at visible light frequencies (atoms)
> > must surely be the most exciting way to study the near and far
> > field anomalies of Maxwell's equations. ( The reason for SR? )
> >
> > For my own personal tastes, I rank the study technique
> > as every bit as exciting as self mutilation, influenza or up
> > close and personal poison plant study.
> >
> > With all those charges and waves whizzing around, I hope
> > I am forgiven for not quite understanding who or what proves
> > or disproves anything.
>
> I was being sarcastic. :-)
LOL I gathered that. I was including myself in the sarcasm at
all of us that spend countless hours shuffling x' and t before
concluding that we'll have to include something different before
we get results any different.
>
> However, you have now given several examples of a disproof of SR,
> or perhaps more properly examples outside of the realm of SR.
> Do they fall outside or disprove GR?
I think it is better to say they are outside the realm of SR at
least as far as SR postulates an observer dependant SOL.
Weber gives some insight to the mechanism at least as far
as interstellar curvature and lensing are concerned.
The whole business of fiddling time instead of energy in
GR makes it difficult, for me at least, to guess what GR's
field equations should represent if they were based on
relativistic electrodynamic field equations instead of
Maxwell's equations.
>
> And, do they fall outside or disprove Newton's "laws of motion"?
AFAIK Newton is silent on the forces between relative moving
charges so woudln't any blarely plausible electrodynamic theory
show Newton's laws to have limitations?
>
> BTW: visible light is not at atomic wavelengths. Visible light
> is in the range of maybe 380 nm to 760 nm. The size of an atom is
> about 100-200 pm. You're talking very hard X-rays, almost
> gamma rays.
Yes, I am sure that could have been better said. The extent
of the electron cloud is generally what I was refering to.
I probably shoud have said the like spectra.
My point was that a sodium atom would not make a very
good television antenna but it works swell for yellow light.
>
> http://mywebpages.comcast.net/bondono2/ispect.html
>
> However, the electron binding energy for, say, neon, is 21.6 eV,
> which translates into the soft X-ray range -- and usually
> one does't even need that much energy, as one need merely
> transition a bound electron (in the case of neon one might
> use a 2p orbital) to another, higher-energy state, then let
> the electron transition back emitting a photon in the process.
Whoa... :-) A copper atom may interacts, but not emit or
absorb the incident EM. It is this aperture to which I was
refering. It can be quite large as the atoms are continually
swaping and sharing outer shell charges.
That is an interesing question you raise about GR's dependance
on SR. I am not sufficiently steeped in GR's field equations to
hazard a guess what would remain if it's imaginary axes were
properly declared but I might look into it.
Sue...
Bilge
>Two particle pairs in relative motion violates the first postulate
>of SR. The magnets would fall off of your refrigerator if it didn't.
Nonsense. You can't even explain the existence of a permanent magnet
without relativity. Feel free to try.
>Two coupled structures in the near-field violate the second
>postulate of SR. The transistiors in your computer would stop
>working if it didn't.
Again, that's nonsense. Try proving that statement.
>The use of visible light at atomic wavelengths coupled to
>structures that change form at visible light frequencies (atoms)
>must surely be the most exciting way to study the near and far
>field anomalies of Maxwell's equations. ( The reason for SR? )
What anomalies?
>For my own personal tastes, I rank the study technique
>as every bit as exciting as self mutilation, influenza or up
>close and personal poison plant study.
Then you should check out trepanning.
>With all those charges and waves whizzing around, I hope
>I am forgiven for not quite understanding who or what proves
>or disproves anything.
Not understanding is one thing. Your cluttering up threads with
unfounded garbage in an attempt to propagate misinformation
to unwary readers who might want to understand is quite another.
eleaticus
"Mich" <mi...@efni.com> wrote in message
news:1135und...@corp.supernews.com...
> As a layman I have some questions, if you don't mind.
> Thanks in advance for your answers.
>
> The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in
> message news:k7kvf2-...@sirius.athghost7038suus.net...
> My first question would be; if c and v are velocity vectors as opposed to
> speed,since it can either be + or -, why isn't x a displacement as opposed
> to a distance, which ought to mean that if x = -ct, then, -x = ct?
They really are vectors.
What discipline's jargon says a linear displacement isn't a distance?
x and x' - outside of physics - are just locations at a particular distance
from the origin, but all rational physical equations in a bare x (as opposed
to something like (x1-x0)) use x as a distance/length of something or from
something, the other end of the distance being at the origin. Hence, they
are distances in physical equations.
eleaticus
eleaticus
> My first question would be; if c and v are velocity vectors as opposed to
> speed,since it can either be + or -, why isn't x a displacement as opposed
> to a distance, which ought to mean that if x = -ct, then, -x = ct?
Pardon me for not immediately recognizing the point behind your question. A
point I've made a number of times in the past.
The expressions x=vt and x=ct do indeed stand for x=(-V)t and x=(+V)t,
x=(-C)t and x=(+C)t. V=|v|, C=|c|.
eleaticus
eleaticus
> As for the x=ct / x=-ct / x=vt controversy, that's probably
> best resolved by using subscripted variables or some such.
ROFFLMFAO!
If he subscripted the two, mutually exclusive x values (except when t=0,
given |c|>0) how would he then bunble his way to a formula for just one x'
in terms of just one x when he also had two x' values?
ROFF:MFAO!
> Einstein, in his thesis (an English transcript is at
> http://www.fourmilab.ch/etexts/einstein/specrel/www/ )
> apparently used x,y,z,t for coordinates in the stationary
> system (termed 'K') and xi (ξ), eta (η),
> zeta (ζ), and tau (τ) for coordinates in the
> moving system (termed 'k').
He then proceeds to use notations
> such as x' = x-vt, tau(x',y,z,t), and partial derivatives,
> ultimately arriving at the result
Best guess is that he kludged at least three previously written papers
together.
And you missed him using x' as a stationary system length.
eleaticus
Bilge
>In sci.physics, jahn
><susyse...@yahoo.com.au>
>> Two particle pairs in relative motion violates the first postulate
>> of SR. The magnets would fall off of your refrigerator if it didn't.
>>
>> Two coupled structures in the near-field violate the second
>> postulate of SR. The transistiors in your computer would stop
>> working if it didn't.
>>
>> The use of visible light at atomic wavelengths coupled to
>> structures that change form at visible light frequencies (atoms)
>> must surely be the most exciting way to study the near and far
>> field anomalies of Maxwell's equations. ( The reason for SR? )
>>
>> For my own personal tastes, I rank the study technique
>> as every bit as exciting as self mutilation, influenza or up
>> close and personal poison plant study.
>>
>> With all those charges and waves whizzing around, I hope
>> I am forgiven for not quite understanding who or what proves
>> or disproves anything.
>
>I was being sarcastic. :-)
>
>However, you have now given several examples of a disproof of SR,
>or perhaps more properly examples outside of the realm of SR.
Those don't fall outside the realm of special relativity. If you
want to understand a permanent magnet, for example, start with the
mass-energy-momentum relations, E^2 = p^2 + m^2 and one supplemental
postulate based on planck's work, namely that E = i\hbar d/dt and
p = -i\hbar\grad. Insert those into the mass energy relation
and you obtain a wave equation, namely the klein-gordon equation
(also called the relativistic shroedinger equation if you ignore
the possibility of spin).
Now, note that you can rewrite the mass energy relation as,
E^2 = (a.p + bm)^2 or, E = +/-(a.p + bm)
which is called the dirac equation. If you solve for the `a_i' and
`b', (since a.p is a ``dot product,'' you have a_x, a_y and a_z
since p has x,y and z components), you get the relations
(a_i)^2 = 1
a_i a_j + a_j a_i = 0 if i != j,
a_i b + b a_i = 0 b^2 = 1
Therefore, the `a_i' and `b' are matrices. From that one finds
that this equation describes particles have an additional, non-
classical degree of freedom which was called ``spin'' and in
particular, has values which are half-integral. So far we have
done nothing but explore more subtle consequences of relativity.
(Since E has both positive and negative solutions, we have also
discovered anti-matter, but that is a differet topic.)
Further below, we'll see that we'll also discover that the
electron has a magnetic moment associated with that spin.
It's conventional to make the relativistic invariance explicit
and multiply through by `b' to obtain,
bE = ba.p + m
and define the dirac matrices, \gamma^u as,
\gamma^0 = b, \gamma^i = ba^i
so that the dirac equation can be written in four-vector notation as,
\gamma^u p_u = m
or, more compactly, p/ = m, where the / indicates the contraction of
p_u with the dirac matrices. The metric can now be written in terms
of the \gamma matrices,
2 g_uv = \gamma^u\gamma^v + \gamma^v\gamma^u
which is the condition for lorentz covariance. We now invoke the
first postulate of special relativity again. Since p/ is a lorentz
scalar and in fact is equal to `m', p/ must be the same for all
observers. In particular, it has to be invariant under lorentz
boosts, space and time displacements and orthogonal rotations,
or in other words, it must be poincare invariant. One can write
the general poincare transform as a unitary transform,
U = \exp(-is), U^-1 = \exp(is) and UU^-1 = U^-1 U = 1
which is most general if one allows `s' itself to be a function of
the coordinates, i.e., s == s(x^u). The transform of p^u from the
unprimed to the primed frame is,
p_u' = U^-1 p_u U
Since U is a function of the variables and p^u = (E, p), which were
defined differential operators, the poincare invariance would appear
to be spoiled and the equation would not satisfy the first postulate
of relativity. After taking derivatives, we have (dropping the irrelevant
\hbar's):
p_u' = \exp(is) p_u \exp(-is) = p_u + d_u s
p_u' is evidently not equal to p_u. But, we can do the following.
Assume that we start with, p_u - id_u s instead and transform that,
p_u' + d_u s' = \exp(is) (p_u + d_u s) \exp(-is)
= p_u + d_u s + \exp(is) (d_u s) \exp(-is)
If p_u' is equal to p_u, then `s' must transform as,
d_u s' = d_u s + d_u s''
where s'' = \exp(is) (d_u s) \exp(-is). We have now discovered that
the original equation requires an ``extra piece.'' That extra piece
is called a ``field,'' and in particular, if we rewrite `s' as
d_u s = eA_u,
then we get, the gauge transformation under which maxwell's equations
transform,
A_u' = A_u + (1/e)d_u s''
such that d_u A^u' = d_u A^u + (1/e)d_u d^u s'' = 0. Special relativity
and the one additional assumption that E and p are given by differential
operators, has required us to include a field, which we see is the
electromagnetic field. We must now rewrite the dirac equation by
including the \gamma^u,
\gamma^u (p_u + eA_u) = (p/ - A/) = m
If we define the \gamma^u as the electromagnetic currrent, j^u, then,
A/ = j_u A^u
Evidently, we have obtained almost all of QED from invariance. All
that's missing is to allow for the possibility that A_u can exist
independently of any charges (i.e., we have free photons) so that
we add the kinetic energy of any free photons to the theory,
(p/ - m) - j.A + (1/4)F^uv F_uv
is the complete QED lagrangian. That's a fairly major accomplishment
considering that the initial requirements we imposed were very
simple and somewhat obvious (e.g., the first postulate could be
regarded as common sense - we expect experiments to give the same
results independent of where and when we do them).
>Do they fall outside or disprove GR?
Let's compare what we just did to general relativity. In general
relativity, the riemann tensor may be obtained from,
[D_u, D_v] V^a = R^a_buv V^b
Where D_u and D_v are the covariant derivatives defined by,
D_u V^a == d_u V^a + C^a_ub V^b, C^a_ub is a christoffel symbol
So, the commutator gives the curvature from which the field equations
(bianchi identities) and equations of motion (geodesic equation) may
be derived. In the example above, the gauge covariant derivative
is given by,
D_u = d_u - ieA_u
and our commutator is,
[D_u, D_v] = (1/ie) F_uv
F_uv is just the maxwell tensor. Let's call that the ``electromagnetic
curvature''. By taking the partial derivative, we get,
d_u F^uv = j^v
Which are just the inhomogeneous maxwell equations, i.e., the equations
of motion for E&M. The homogeneous equations are the equivalent of
the bianchi identities:
d^a F^bc + d^c F^ab + d^b F^ca = 0
The christoffel symbols are the connection coefficients in general
relativity. In E&M the connection coefficient is iA_u. The presence
of the `i' prevents us from associating the connection with the
curvature of our four-dimensional geometry. We can instead associate
that with a fiber bundle or a ``curled up dimension'' (ala string
theory). The fiber bundle just attaches a U(1) bundle (symmetry group)
to each point in the spacetime.
The magnetic moment of the electron may be obtained by solving for
the electromagnetic current, j^u. Since j^u = \gamma^u and the
spin associated with the pauli matrices is the non-relativistic
reduction, we have,
\gamma^u p_u = m
which we multiply by \gamma^v to obtain,
\gamma^v\gamma^u p_u = \gamma^v m
or,
\gamma^v = (1/m)\gamma^v\gamma^u p_u
From the identity given earlier,
g_uv = (1/2) (\gamma^u\gamma^v + \gamma^v\gamma^u)
and a second definition,
\sigma_uv = (i/2) (\gamma^u\gamma^v - \gamma^v\gamma^u)
We can write \gamma^v\gamma^u as,
\gamma^v\gamma^v = g^uv + i\sigma^uv
so that if \gamma^v is given by the expression above,
\gamma^v = g^vu p_u + i\sigma^vu p_u
If you evaluate this as the electromagnetic interaction between
some initial state and final state, you see it has two terms.
The first is just the usual momentum so that it will be proportional
to only the magnitude of the charge. The second term contains
the spin, which implies the charge also acts through the spin of
the particle, hence the magnetic moment. Note that the effect has
nothing to do with rotation. It's a consequence of special relativity.
So, if you now want to apply that to a permanent magnet, you
see that (1) there are two parts to the electromagnetic current,
(2) the part due to moving charges is absent, (3) there is still
a current due to the electron spin, (4) if the potential minimum
is lowest when the spins are aligned, the current due to the spins
is responsible for an overall magnetization, (5) we didn't have
to even assume E&M to obtain any of this - just relativity plus
planck's results.
>And, do they fall outside or disprove Newton's "laws of motion"?
Newton's laws will remain forever as the low velocity, macroscopic
physics of this universe. They are only ``disproved'' to the extent
that they fail to describe phenomena for which they were never
conceived, as newton knew nothing of the phenomena for which his
laws require modification.
The Ghost In The Machine
<39gto4F...@individual.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:1kmbg2-...@sirius.athghost7038suus.net...
[snip stuff for brevity]
>
> That is an interesing question you raise about GR's dependance
> on SR.
Dependence? I'm not sure that's the right word. SR is GR
with zero gravity/strain-free space, as I understand it.
Certainly GR can deal with most if not all of the messy stuff
that SR skips around (acceleration, geodesics, and gravity).
> I am not sufficiently steeped in GR's field equations to
> hazard a guess what would remain if it's imaginary axes were
> properly declared but I might look into it.
Imaginary axes? What imaginary axes?
Granted, one can set up either an 8-space over R, a 4-space
over C, or a 4-space over R with an unusual measurement metric
(d^2 = (x-x0)^2 + (y-y0)^2 + (z-z0)^2 + (ict-ict0)^2).
The Ghost In The Machine
<dub...@radioactivex.lebesque-al.net>
wrote
on Sat, 12 Mar 2005 22:42:54 GMT
<slrnd3728o....@radioactivex.lebesque-al.net>:
> jahn:
>
> >Two particle pairs in relative motion violates the first postulate
> >of SR. The magnets would fall off of your refrigerator if it didn't.
>
> Nonsense. You can't even explain the existence of a permanent magnet
> without relativity. Feel free to try.
OK, dumb question. How does one explain the existence of such
a magnet *with* relativity? :-) The best I can do is related to
the issue of brehmsstrahlung, and that's more covered by QM
than SR/GR.
(Briefly, brehmsstrahlung is theoretical radiation by accelerating
charges; were we in a Newtonian space, electrons would eventually
spiral into the proton. Since they obviously don't, Newton is
crap at this level. This also appears related to the ultraviolet
catastrophe.)
>
> >Two coupled structures in the near-field violate the second
> >postulate of SR. The transistiors in your computer would stop
> >working if it didn't.
>
> Again, that's nonsense. Try proving that statement.
>
> >The use of visible light at atomic wavelengths coupled to
> >structures that change form at visible light frequencies (atoms)
> >must surely be the most exciting way to study the near and far
> >field anomalies of Maxwell's equations. ( The reason for SR? )
>
> What anomalies?
I wonder that myself. Compton's effect will probably have to
deal with such anomalies.
>
> >For my own personal tastes, I rank the study technique
> >as every bit as exciting as self mutilation, influenza or up
> >close and personal poison plant study.
>
> Then you should check out trepanning.
But that's just boring! :-) Or are you trying to trick her? :-)
[rest snipped]
The Ghost In The Machine
<elea...@bellsouth.net>
wrote
on Sat, 12 Mar 2005 18:42:42 -0600
<f4MYd.26507$5T6....@bignews4.bellsouth.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:7hlbg2-...@sirius.athghost7038suus.net...
>
>> As for the x=ct / x=-ct / x=vt controversy, that's probably
>> best resolved by using subscripted variables or some such.
>
> ROFFLMFAO!
>
> If he subscripted the two, mutually exclusive x values (except when t=0,
> given |c|>0) how would he then bunble his way to a formula for just one x'
> in terms of just one x when he also had two x' values?
>
> ROFF:MFAO!
Tsk tsk.
Let's set it up properly then.
Let (x,y,z,t) be one coordinate system, and (X,Y,Z,T) be another.
These coordinate systems are related, hypothetically, by the
following:
[1] If the equation x=vt+k holds true, then X=K in the other system,
for some value of K depending on v and k.
[2] If the equation X=-vT+K holds true, then x=k in the other system,
for some value of k depending on v and K.
[3] If the equation x=ct holds true, then X=cT in the other system.
[4] The transformation between (x,y,z,t) and (X,Y,T,Z) is linear.
Now...is there a problem with this setup? Should I use set theory,
instead? Assuming LT(v) is a tensor -- a transformation of one
point of 4-space into another point of 4-space, in this case --
and P * LT(v) is the transformation of point P by the tensor LT(v),
we might have the following.
[1'] If one has the set S={(vt+k,0,0,t): t in R},
then S' = S * LT(v) has S'={(K,0,0,T): T in R}
for some K = K(v,k).
[2'] If one has the set S'={(-vT+K,0,0,T): T in R},
then S = S' * LT(-v) has S={(k,0,0,t): t in R}
for some k = k(v,K).
[3'] The set S={(ct,0,0,t): t in R} is such that S = S * LT(v).
In other words, S transforms to itself.
[4'] ((a,b,c,d) + (e,f,g,h)) * LT(v)
= (a,b,c,d) * LT(v) + (e,f,g,h) * LT(v)
for all a,b,c,d,e,f,g,h in R.
The most problematic assumption is arguably [3] or [3'], mostly
because it's outside of most humans' experience. I stand
on top of a platform moving at 10 m/s (don't try this without
sufficient protection) and throw a ball at 1 m/s. The ball
moves at 11 m/s relative to the ground, right? Well, close.
The SR velocity differs from 11 m/s by about 1.223915 * 10^-15 m/s.
This is probably not something that's going to be easy to measure.
But so far, *everyone* has measured light speed to be c = 299792458 m/s.
In fact, it's now defined that way.
I'm also curious as to why you restrict t > 0. Is there a reason?
>
>> Einstein, in his thesis (an English transcript is at
>> http://www.fourmilab.ch/etexts/einstein/specrel/www/ )
>> apparently used x,y,z,t for coordinates in the stationary
>> system (termed 'K') and xi (ξ), eta (η),
>> zeta (ζ), and tau (τ) for coordinates in the
>> moving system (termed 'k').
> He then proceeds to use notations
>> such as x' = x-vt, tau(x',y,z,t), and partial derivatives,
>> ultimately arriving at the result
>
> Best guess is that he kludged at least three previously written papers
> together.
>
> And you missed him using x' as a stationary system length.
>
> eleaticus
>
Perhaps you would still like to offer an alternative theory, one that:
[1] mathematically works?
[2] illustrates and predicts all known relevant phenomena, such as
Mercury's perturbations and the Compton effect?
eleaticus
> >> As for the x=ct / x=-ct / x=vt controversy, that's probably
> >> best resolved by using subscripted variables or some such.
> > ROFFLMFAO!
> > If he subscripted the two, mutually exclusive x values (except when t=0,
> > given |c|>0) how would he then bunble his way to a formula for just one
x'
> > in terms of just one x when he also had two x' values?
> > ROFFLMFAO!
> Tsk tsk.
> Let's set it up properly then.
OK. are you or are you not 'rebutting' me?
You say to use subscripted variables - which would be nonsense if you wren't
talking about one x for ct and another for -ct and then you don't?
LMFAOWYMAAOY!
> Let (x,y,z,t) be one coordinate system, and (X,Y,Z,T) be another.
> These coordinate systems are related, hypothetically, by the
> following:
>
> [1] If the equation x=vt+k holds true, then X=K in the other system,
> for some value of K depending on v and k.
> [2] If the equation X=-vT+K holds true, then x=k in the other system,
> for some value of k depending on v and K.
> [3] If the equation x=ct holds true, then X=cT in the other system.
> [4] The transformation between (x,y,z,t) and (X,Y,T,Z) is linear.
> Now...is there a problem with this setup?
Yes, of course. You insisted it was ok to use x=ct, x=-ct, and x=vt
simultaneously?
What happened to that fine, marvellous usage?
eleaticus
Sam Wormley
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:7hlbg2-...@sirius.athghost7038suus.net...
>
>
>>As for the x=ct / x=-ct / x=vt controversy, that's probably
>>best resolved by using subscripted variables or some such.
>
>
> ROFFLMFAO!
>
YLFWYAO!
FrediFizzx
| In sci.physics, Bilge
| <dub...@radioactivex.lebesque-al.net>
| wrote
| on Sat, 12 Mar 2005 22:42:54 GMT
| <slrnd3728o....@radioactivex.lebesque-al.net>:
| > jahn:
| >
| > >Two particle pairs in relative motion violates the first postulate
| > >of SR. The magnets would fall off of your refrigerator if it
didn't.
| >
| > Nonsense. You can't even explain the existence of a permanent
magnet
| > without relativity. Feel free to try.
|
| OK, dumb question. How does one explain the existence of such
| a magnet *with* relativity? :-) The best I can do is related to
| the issue of brehmsstrahlung, and that's more covered by QM
| than SR/GR.
Didn't you bother to follow the derivation Bilge posted? Here is
something simpler. magnetic charge would be proportional to e*c.
Magnetism is a relativistic effect.
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
Bilge
><dub...@radioactivex.lebesque-al.net>
> wrote
>on Sat, 12 Mar 2005 22:42:54 GMT
><slrnd3728o....@radioactivex.lebesque-al.net>:
>> jahn:
>>
>> >Two particle pairs in relative motion violates the first postulate
>> >of SR. The magnets would fall off of your refrigerator if it didn't.
>>
>> Nonsense. You can't even explain the existence of a permanent magnet
>> without relativity. Feel free to try.
>
>OK, dumb question. How does one explain the existence of such
>a magnet *with* relativity? :-) The best I can do is related to
>the issue of brehmsstrahlung, and that's more covered by QM
>than SR/GR.
line is that the electromagnetic current can be separated into two
components. One part is the interaction one usually associates with
the classical coulomb force. The other part is the interaction through
the electron spin. The spin is a relativistic feature. The current
responsible for the magnetization of a permanent magnet is the part
due to the electron spin.
>(Briefly, brehmsstrahlung is theoretical radiation by accelerating
>charges; were we in a Newtonian space, electrons would eventually
>spiral into the proton. Since they obviously don't, Newton is
>crap at this level. This also appears related to the ultraviolet
>catastrophe.)
>
>>
>> >Two coupled structures in the near-field violate the second
>> >postulate of SR. The transistiors in your computer would stop
>> >working if it didn't.
>>
>> Again, that's nonsense. Try proving that statement.
>>
>> >The use of visible light at atomic wavelengths coupled to
>> >structures that change form at visible light frequencies (atoms)
>> >must surely be the most exciting way to study the near and far
>> >field anomalies of Maxwell's equations. ( The reason for SR? )
>>
>> What anomalies?
>
>I wonder that myself. Compton's effect will probably have to
>deal with such anomalies.
>
>>
>> >For my own personal tastes, I rank the study technique
>> >as every bit as exciting as self mutilation, influenza or up
>> >close and personal poison plant study.
>>
>> Then you should check out trepanning.
>
>But that's just boring! :-) Or are you trying to trick her? :-)
There's no pun like a bad pun...
The Ghost In The Machine
<slrnd37mpo....@radioactivex.lebesque-al.net>:
> The Ghost In The Machine:
> >In sci.physics, Bilge
> ><dub...@radioactivex.lebesque-al.net>
> > wrote
> >on Sat, 12 Mar 2005 22:42:54 GMT
> ><slrnd3728o....@radioactivex.lebesque-al.net>:
> >> jahn:
> >>
> >> >Two particle pairs in relative motion violates the first postulate
> >> >of SR. The magnets would fall off of your refrigerator if it didn't.
> >>
> >> Nonsense. You can't even explain the existence of a permanent magnet
> >> without relativity. Feel free to try.
> >
> >OK, dumb question. How does one explain the existence of such
> >a magnet *with* relativity? :-) The best I can do is related to
> >the issue of brehmsstrahlung, and that's more covered by QM
> >than SR/GR.
>
> It's not a dumb question. See my other post for details.
Ah, well, I'd not actually seen your other post prior
to my posting the above one. I shall have to study
it; electrodynamics regrettably is one subject I
skipped in college.
Not that it matters all that much; it merely means that
poor jahn/sue is confused again. :-) Or perhaps
trying to confuse us.
In any event, Einstein's 1905 paper is titled
ON THE ELECTRODYNAMICS OF MOVING BODIES
which presumably means electromagnetism gets in there
at some point -- and it does, in Part II.
Oh yes, they're the best kind. :-) Or perhaps the worst.
I'll have to remember that one as I pan for more outré puns
myself...
The Ghost In The Machine
<nfOYd.25581$Q83....@bignews5.bellsouth.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in
> message news:5flcg2-...@sirius.athghost7038suus.net...
>
>> >> As for the x=ct / x=-ct / x=vt controversy, that's probably
>> >> best resolved by using subscripted variables or some such.
>
>> > ROFFLMFAO!
>
>> > If he subscripted the two, mutually exclusive x values (except when t=0,
>> > given |c|>0) how would he then bunble his way to a formula for just one
> x'
>> > in terms of just one x when he also had two x' values?
>
>> > ROFFLMFAO!
>
>> Tsk tsk.
>
>> Let's set it up properly then.
>
> OK. are you or are you not 'rebutting' me?
Apparently I'm not rebutting you since you're either
falling off your chair and injuring myself (fortunately
there are many fine medical plans available, if one is
gainfully employed), or completely ignoring the facts
(mostly of the form "these experiments seem to be producing
very consistent results with the theory as presented") in
order to focus on trivial issues such as variable scoping,
and laughing your head off. While somewhat therapeutic
(presumably because phlegm is expelled from the lungs,
although an asthmatic attack is also likely), it doesn't
seem to be entirely in the realm of physics.
>
> You say to use subscripted variables - which would be nonsense if you wren't
> talking about one x for ct and another for -ct and then you don't?
>
> LMFAOWYMAAOY!
Ah, well, apparently I'm not making myself sufficiently clear.
>
>
>> Let (x,y,z,t) be one coordinate system, and (X,Y,Z,T) be another.
>> These coordinate systems are related, hypothetically, by the
>> following:
>>
>> [1] If the equation x=vt+k holds true, then X=K in the other system,
>> for some value of K depending on v and k.
>> [2] If the equation X=-vT+K holds true, then x=k in the other system,
>> for some value of k depending on v and K.
>> [3] If the equation x=ct holds true, then X=cT in the other system.
>> [4] The transformation between (x,y,z,t) and (X,Y,T,Z) is linear.
>
>> Now...is there a problem with this setup?
>
> Yes, of course. You insisted it was ok to use x=ct, x=-ct, and x=vt
> simultaneously?
It is if one scopes it properly.
S_2={x^2: x in R}
S_4={x^4: x in R}
Are these sets referring to the same x? Yes? No?
It turns out these two sets are equal (S_2=S_4=R+ union {0}), just
to make things weirder.
>
> What happened to that fine, marvellous usage?
Oh, you would rather focus on the details than the grand overall
scheme. Fine.
Let P_a = (x_a, y_a, z_a, t_a) be a 4-dimensional co-ordinate system
for the observer A, and P_b = (x_b, y_b, z_b, t_b) be a 4-dimensional
co-ordinate system for the observer B.
Let B be moving uniformly (no acceleration) relative to A, and
let A and B be coincident at some point. Define that point
in A-space as (0,0,0,0), and in B-space also as (0,0,0,0), by
moving the origins in both A and B space as necessary (we assume
non-distorted space). Furthermore, assume that B is moving along
A'x x_a-axis, and that A is moving (in the opposite direction) along
B's x_b-axis, by rotating the co-ordinate systems as necessary.
We notate the transformation from A-space to B-space by using
P_b = P_a * LT(v)
where LT(v) is a function with domain R (since v is a scalar, not a
vector, in this particular problem) and a range which is a little
hard to describe unless one throws in the word tensor, but
basically the range is a tensor from one 4-space to another.
The following are also assumed.
[1] If x_a = v * t_a, then the transformed x_b = 0. This is because
the point in A-space is assumed coincident with B's origin.
(Note that the scope of this equality does NOT apply outside
of this clause. If you want to quibble about this I can introduce
other notations such as S_aV = {(v * t_a_aV, 0,0,t_a_aV): t_a_aV in R}
and S_ac = {(c * t_a_ac, 0,0,t_a_ac): t_a_ac in R}, but that gets
rather unwieldly.)
[2] Similarly, if x_b = -v * t_b, then x_a = 0.
[3] If x_a = c * t_a + x_a0 for some x_a0, where c > 0 is the speed
of light, then after transformation x_b = c * t_b + x_b0
for some x_b0 which depends on x_a0, v, and c.
[4] The transformation from P_a to P_b is linear. That is to say,
if one has two points P_a and Q_a, then it must be the case
that
(P_a + Q_a) * LV(a) = (P_a * LT(v)) + (Q_a * LT(v))
[5] LT(-v) = LT(v)^-1.
Using these simplified assumptions, the Lorentz falls in one's lap,
with a little work.
Now, is this clear yet or should I tackle your funny bone some more?
[rest snipped]
jahn
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:u4kcg2-...@sirius.athghost7038suus.net...
> In sci.physics, jahn
> <susyse...@yahoo.com.au>
> wrote
> on Sat, 12 Mar 2005 14:19:55 -0500
> <39gto4F...@individual.net>:
> >
> > "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message
news:1kmbg2-...@sirius.athghost7038suus.net...
>
> [snip stuff for brevity]
>
> >
> > That is an interesing question you raise about GR's dependance
> > on SR.
>
> Dependence? I'm not sure that's the right word.
SR lays a foundation for imaginary time axes.It's not a hard
thing to do since time is such a abstraction anyway. Is is a
is a swell way to account for mass and energy that is spatially
dispersed and it conviently hides the near-field anomalies
in Maxwell's equaitons.
The sin is, faiure to describe the new axes and subsequently
treating them as real.
>SR is GR
with zero gravity/strain-free space, as I understand it.
> Certainly GR can deal with most if not all of the messy stuff
> that SR skips around (acceleration, geodesics, and gravity).
Hmmm... In the sense that Maxwell and Einstein's field equation
have some commonality in their evolution that may be true.
But I am don't know where Maxwell ever suggested that a
coordinate system moving wrt a particle could produce a force.
>
> > I am not sufficiently steeped in GR's field equations to
> > hazard a guess what would remain if it's imaginary axes were
> > properly declared but I might look into it.
>
> Imaginary axes? What imaginary axes?
Exactly! Had AE properly declared them you would know.
The complex (imaginary) wave impedance changes in the
near field of a coupled structure, causing the path to be
inhomnogenous. Dumping these effects into a remotly viewed
clock and claiming mastry over time might be clever but the
twins gedanken bears testiment that the approach has it's
limitations.
Sue...
Mich
greater details my problem.
The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in
message news:7hlbg2-...@sirius.athghost7038suus.net...
> In sci.physics, Mich
> <mi...@efni.com>
> wrote
> on Sat, 12 Mar 2005 09:24:22 -0500
> <1135und...@corp.supernews.com>:
> >
> > The Ghost In The Machine <ew...@sirius.athghost7038suus.net> wrote in
> > message news:k7kvf2-...@sirius.athghost7038suus.net...
> >> In sci.physics, eleaticus
> >> <elea...@bellsouth.net>
> >
> >> > -----------------------------------
> >> > Einstein initiates his process by asserting four basic equations and
> >> > adds a fifth in his process:
> >> >
> >> > x = ct
> >> > x = -ct
> >> > x = vt
> >> >
> >> > x' = ct'
> >> > x' = -ct'
> >>
> >> OK, we have x, t, x', t', as variables, and c and v as constants.
> All values are scalars, though if you like one can move into the
> vector realm by replacing x with X or P (don't use 'p'; that's too
> easily confused with momentum :-) ).
I think that I understand x to be a scalar if it is to mean a distance,
identifying a negative
distance as being illogical. However, if x is a product of a velocity, not
speed, then it ought to be specified as a displacement, and not only a
distance, it seems.
The + or - sign in front of a velocity vector simply indicates the
direction of travel,and could therefore simply mean east or west. The same
could be said for the displacement, where -x could mean the point at
distance x (west) from the observer and + x meaning the point at distance x
(east) from the observer.
So in the above equations, I would it understand better if x = ct and -x
= -ct. As it stands, I don't understand the meaning of the equation.
I will cut the rest of your post so as not to venture too quickly before
understanding the basics.
Thank you for your reply
Andre
.
Mich
A
> point I've made a number of times in the past.
>
> The expressions x=vt and x=ct do indeed stand for x=(-V)t and x=(+V)t,
> x=(-C)t and x=(+C)t. V=|v|, C=|c|.
O.K You might have the answer I'm looking for;
In the equations below
x = ct
x = -ct
x = vt
Why is ( x = -ct) not (- x = -ct) instead ?
>
> eleaticus
Andre
eleaticus
Both of Einstein's derivations involve logical absurdities of this type that
would make his x'=g(x-vt) and t'=g(t-vx/cc) valid only when t=0, at most.
-x = -ct is the same as x=ct and thus provides no grounds for eventually
concluding absurdly with the above transformations.
Only with absurd input can you reach absurd consequences.
Do not suppose that Einstein's work in this matter can be explained
logically. His x, by the way, was whatever vt and ct and -ct made it, and
his v was |v| < c.
Did you see my tale about Dirk (Dimmy)?
Here is a repeat of part of a recent post in response to fantasies:
----------------------------------------------
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet the only equation (plus dengenerate versions) that is
actually invariant under Einstein's xforms is x^2+y^2+z^2-(ct)^2.
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet x=ct and x=-ct have common solutions only where c=0 or t=0.
The transforms are supposed to be valid for all values of x, any t>=0, and
any |v|<|c|, yet x=ct and x=vt have common solutions only where v=c or t=0.
Soon after Dirk The Dim's 33rd birthday his Mommy decided to let the lad
take a trip on his own. A family friend owned a hotel in a nearby town and
promised to look after the boy.
Happily, Dimmy arrived at the hotel but soon was crestfallen. He had
forgotten his pencils. Oh, shoot! He had so terribly wanted to figure out
whether or not it was really true that 3+2=5.
He asked the hotel desk clerk if there was any place nearby where he could
buy pencils. At first the clerk was reluctant to answer but he checked with
the owner and was told it was ok to tell the lad where it could be done, but
the clerk would have to chaperon the boy.
"Go out the front doors and turn left, Dimmy, and one block away you can buy
pencils at only $1.20 a dozen at the convenience store."
"Ohh!" exclaimed the lad, "that will kill my allowance!"
"Or, you can turn right and buy pencils for $1.25 a dozen at the drugstore."
Dimmy decided a drugstore's pencils would surely be worth the extra nickle a
dozen so the clerk took him by the hand and helped him get there and back.
Dimmy so so overwhelmed by the adventure that he couldn't concentrate on
what he had thought was a major problem, is 2+3 equal to 5.
What, he wondered, is the mathematical law governing pencil prices in this
town? Surely there must be one.
Let the total cost C be 1.20D in the one direction, he thought, and 1.25D in
the other direction.
That gives us C=1.20D and C=1.25D, he continued, and 2C = 2.45D, and
C=1.225D.
Wow! he thought, I did it! The underlying law of pencil prices in this town!
Mommy will be so proud!
Little did Dimmy understand, later, why people said you can't use
contradictories simultaneously in a continuous derivation. That their only
use simultaneously was to see if there was a common solution. That any
derivation could only apply to the common values, if any.
Some even said that a whole logical proof technique was based on finding
just one pair of contradictories, reduction to the absurd.
But Dimmy knew better. His only regret was that there hadn't been a third
source for pencils, maybe one for C=1.00D, which would have given him
C=1.15D, a much more elegant and beautiful mathematical law!
> >
> > eleaticus
>
> Andre
>
>
>
Bilge
>
>to my posting the above one. I shall have to study
>it; electrodynamics regrettably is one subject I
>skipped in college.
appear.
>
>Not that it matters all that much; it merely means that
>poor jahn/sue is confused again. :-) Or perhaps
>trying to confuse us.
>
>In any event, Einstein's 1905 paper is titled
>
>ON THE ELECTRODYNAMICS OF MOVING BODIES
>
>which presumably means electromagnetism gets in there
>at some point -- and it does, in Part II.
Einstein's intent was to justify maxwell's equations as having
a basis in geometry. My intent was somewhat different. I wanted to
point out that it's possible to _derive_ maxwell's equations and
some features that aren't present in maxwell's equations, using
little more than relativity and really, just the first postulate.
Dropping the second postulate only entails ending up with a massive
photon using the same basic idea.
Mich
>
> "Mich" <mi...@efni.com> wrote in message
> news:1139f50...@corp.supernews.com...
> >
> > eleaticus <elea...@bellsouth.net> wrote in message
> > news:p_LYd.26480$5T6....@bignews4.bellsouth.net...
> > >
> > > "Mich" <mi...@efni.com> wrote in message
> > > news:1135und...@corp.supernews.com...
> > > Pardon me for not immediately recognizing the point behind your
> question.
> > A
> > > point I've made a number of times in the past.
> > >
> > > The expressions x=vt and x=ct do indeed stand for x=(-V)t and x=(+V)t,
> > > x=(-C)t and x=(+C)t. V=|v|, C=|c|.
> >
> > O.K You might have the answer I'm looking for;
> >
> > In the equations below
> > x = ct
> > x = -ct
> > x = vt
> >
> > Why is ( x = -ct) not (- x = -ct) instead ?
>
> Both of Einstein's derivations involve logical absurdities of this type
that
> would make his x'=g(x-vt) and t'=g(t-vx/cc) valid only when t=0, at most.
Since I am in the process in trying to understand each steps involved, I
will refrain from commenting,
and will cut everything else for the purpose of not skipping any of the
steps involved. But this does not mean that I haven't read all of the post.
>
> -x = -ct is the same as x=ct and thus provides no grounds for eventually
> concluding absurdly with the above transformations.
Ok...why are these two statements the same? In my mind, I thought x meant a
point at X distance from the observer in the direction - ( which could mean
North, for example), while + x would mean point, the same distance from the
observer, but in the opposite direction(South).
If that's the case, then, I don't see how both equations can mean the same
thing.
> > > eleaticus
Andre
>
Tom Capizzi
>
> eleaticus <elea...@bellsouth.net> wrote in message
> news:lF7Zd.26038$6g7....@bignews1.bellsouth.net...
>>
>> "Mich" <mi...@efni.com> wrote in message
>> news:1139f50...@corp.supernews.com...
>> >
>> > eleaticus <elea...@bellsouth.net> wrote in message
>> > news:p_LYd.26480$5T6....@bignews4.bellsouth.net...
>> > >
>> > > "Mich" <mi...@efni.com> wrote in message
>> > > news:1135und...@corp.supernews.com...
>> > > Pardon me for not immediately recognizing the point behind your
>> question.
>> > A
>> > > point I've made a number of times in the past.
>> > >
>> > > The expressions x=vt and x=ct do indeed stand for x=(-V)t and
>> > > x=(+V)t,
>> > > x=(-C)t and x=(+C)t. V=|v|, C=|c|.
>> >
>> > O.K You might have the answer I'm looking for;
>> >
>> > In the equations below
>> > x = ct
>> > x = -ct
>> > x = vt
>> >
>> > Why is ( x = -ct) not (- x = -ct) instead ?
>>
>> Both of Einstein's derivations involve logical absurdities of this type
> that
>> would make his x'=g(x-vt) and t'=g(t-vx/cc) valid only when t=0, at most.
To eleaticus:
Why don't you cite exactly where in Einstein's derivation this alleged
"absurdity" takes place? Afraid someone will show you your error?
>
> Since I am in the process in trying to understand each steps involved, I
> will refrain from commenting,
> and will cut everything else for the purpose of not skipping any of the
> steps involved. But this does not mean that I haven't read all of the
> post.
>
>>
>> -x = -ct is the same as x=ct and thus provides no grounds for eventually
>> concluding absurdly with the above transformations.
>
> Ok...why are these two statements the same? In my mind, I thought x meant
> a
> point at X distance from the observer in the direction - ( which could
> mean
> North, for example), while + x would mean point, the same distance from
> the
> observer, but in the opposite direction(South).
> If that's the case, then, I don't see how both equations can mean the same
> thing.
To Andre:
It's a simple mathematics principle. Whether x is a distance or a dimension
is
irrelevant. All that matters is that it means exactly the same thing on both
sides
of the equation. Same for c and t. Given an equation, you can multiply both
sides by any non-zero scalar without changing the meaning of the equation.
In this case, if -x is opposite to x, so is -c opposite to c. Technically,
the minus
sign could also be associated with t instead of c, but the combination of c
and
(-t) is still opposite x, or equal to (-x).
From another point of view, you could assume that the two equations were not
the same. Given that c is some constant, you would then have two equations
in two variables. However, if you attempt to eliminate one variable by
adding
the two equations together, the result is 0 = 0, which contains no
information
about the second variable. Hence, the system of equations is insoluble, and
the original asumption that the two equations were not the same must be
false.
>
>
>> > > eleaticus
>
> Andre
>>
>
>
Mich
Tom Capizzi <etian...@verizon.net> wrote in message
news:xhhZd.2922$db6.144@trndny02...
>...
I do agree with you,Tom, that it's a simple mathematical procedure; I was
thinking more in the line of whether it would be legal to use it in such a
case?
If I rewrite the equation as such:
Instead of x = ct, I'd write x(east) = c(east) t
And for x = -ct , I'd write instead x(east)= c(west) t. In my opinion,
since the light starts from the source point 0 towards a direction [c(west)]
opposite than the coordinate point identified [x(east)], the statement
doesn't seem to make any sense, in my opinion.
If we say that the light starts from source point 0,and travels at
velocity(not speed) c,then, at time t, it will have reach coordinate point
x. Also, if I say that the light starts at point 0, and travels at
velocity -c, it must reach coordinate point -x, at time t.
Now, from this, to claim that the two statements -x = -ct, and x = ct are
the same would be true only if we speak in terms of scalars(the distances
the light will have reached at time t will be the same in both cases),
however, if we speak of vectors, the displacement result isn't the same.
Andre
The Ghost In The Machine
<etian...@verizon.net>
wrote
on Mon, 14 Mar 2005 14:34:05 GMT
<xhhZd.2922$db6.144@trndny02>:
I for one think part of the problem may be that in Einstein's work
(or at least this one transcription I have thereof), he used
(x,y,z,t) for one coordinate system, and (xi, eta, zeta, tau)
for the other, and set x' = x-vt. However, most people here,
because of ASCII limitations, use (x',y',z',t') for the other
coordinate system, causing some confusion unless very carefully
handled.
I'll admit I do not know where there is a scan of the original, but
http://www.fourmilab.ch/etexts/einstein/specrel/www/
is an English translation (presumably the original was in German).
The footnotes are interesting, pointing to a number of minor
corrections.
The transcription using beta where today we'd use gamma doesn't
exactly help, either.
Dirk Van de moortel
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:d2ogg2...@sirius.athghost7038suus.net...
This is a scan of the original:
http://www.wiley-vch.de/berlin/journals/adp/890_921.pdf
and this is a crippled transcription of it:
http://www.physics.utoledo.edu/~ljc/speciaal.html
(the equations are missing, but the text is "copy/pastabable")
Dirk Vdm
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
on Mon, 14 Mar 2005 16:50:27 GMT
<nhjZd.1609$mV7...@news.cpqcorp.net>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:d2ogg2...@sirius.athghost7038suus.net...
[snip]
>> I'll admit I do not know where there is a scan of the original, but
>> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>> is an English translation (presumably the original was in German).
>
> This is a scan of the original:
> http://www.wiley-vch.de/berlin/journals/adp/890_921.pdf
> and this is a crippled transcription of it:
> http://www.physics.utoledo.edu/~ljc/speciaal.html
> (the equations are missing, but the text is "copy/pastabable")
Ah, interesting; thanks. :-) This scan clearly shows Einstein's
use of xi, eta, zeta, and tau; even one who can' t read a scrap
of German can see that. (Me, my German's extremely rusty, but
the math is clear, even if the scan pixellated the document.)
Of course I'm half an idiot because the first link is readily
available in the document I show. Doh!
Oh well. :-)
[snip]
eleaticus
> > -x = -ct is the same as x=ct and thus provides no grounds for eventually
> > concluding absurdly with the above transformations.
> Ok...why are these two statements the same? In my mind, I thought x meant
a
> point at X distance from the observer in the direction - ( which could
mean
> North, for example), while + x would mean point, the same distance from
the
> observer, but in the opposite direction(South).
> If that's the case, then, I don't see how both equations can mean the same
> thing.
Hi, Mich!
If -a = -b then a = b. Saying one says the other.
It's the relationships among the terms that count, not the superficial
differences that appear in the equations.
a+c = b+c => a + b.
Neither of those two superficially different equation sets allows absurd
algebraicly correct results.
But x = -ct and x = ct does.
For instance multiply side by side:
x^2 = -(ct)^2
x= ict.
Where did i (imaginary root of -1) come from? It isn't implied or explicit
in either original equation; x,c,t are all real numbers. It is the result of
the absurd contradictions inherent in the general simultaneous use of those
equations. Yes, it too has a valid solution where t=0, but, again, Einstein
was supposed to be deriving equations for all t>0, all x.
As we have seen, just one equation, x=ct (x=-ct would also work) covers all
values of x or t, for -c or +c.
No absurd result algebraicly possible.
eleaticus
eleaticus
> Now, from this, to claim that the two statements -x = -ct, and x = ct
are
> the same would be true only if we speak in terms of scalars(the distances
> the light will have reached at time t will be the same in both cases),
> however, if we speak of vectors, the displacement result isn't the same.
x is by definition not a scalar, it being a coordinate and hence must change
sign as the coordinate axis switches + and - directions.
However, a scalar distance D would suffer even more, but more obviously.
D=ct and D=-ct would make even SR-cult cretins sit up and think: "Oops! That
can't be for c>0 unless t=0!"
eleaticus
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:iAlZd.17664$c72....@bignews3.bellsouth.net...
Congratulations. You have proven analytic geometry wrong again.
Draw two lines on graphing paper:
Line 1: y = 7
Line 2: y = 3
Substract side by side:
0 = 4
Where did 4 come from?
You really are enormously stupid, don't you agree?
Dirk Vdm
Dirk Van de moortel
"Mich" <mi...@efni.com> wrote in message news:113ba73...@corp.supernews.com...
[snip]
>
> I do agree with you,Tom, that it's a simple mathematical procedure; I was
> thinking more in the line of whether it would be legal to use it in such a
> case?
> If I rewrite the equation as such:
> Instead of x = ct, I'd write x(east) = c(east) t
> And for x = -ct , I'd write instead x(east)= c(west) t. In my opinion,
> since the light starts from the source point 0 towards a direction [c(west)]
> opposite than the coordinate point identified [x(east)], the statement
> doesn't seem to make any sense, in my opinion.
> If we say that the light starts from source point 0,and travels at
> velocity(not speed) c,then, at time t, it will have reach coordinate point
> x. Also, if I say that the light starts at point 0, and travels at
> velocity -c, it must reach coordinate point -x, at time t.
> Now, from this, to claim that the two statements -x = -ct, and x = ct are
> the same would be true only if we speak in terms of scalars(the distances
> the light will have reached at time t will be the same in both cases),
> however, if we speak of vectors, the displacement result isn't the same.
>
>
> Andre
Supposing c > 0,
when you write x = c t , you describe a light signal
going in the positive x direction, that happened to be
at position x = 0 when your time read t = 0.
The speed of the signal is c and the velocity is c as well.
At time -T, the signal was at position x = - c T
At time 0, the signal was at position x = 0
At time T, the signal is at position x = c T.
In stead of writing
x = c t
you can just as well write
-x = -c t, or
-12 x = - 12 c t, or
exp(10 x) = exp(10 c t)
Still supposing c > 0,
when you write x = - c t , you describe a light signal
going in the negative x direction, that happened to be
at position x = 0 when your time read t = 0.
The speed of the signal is c but the velocity is -c.
At time -T, the signal was at position x = c T
At time 0, the signal was at position x = 0
At time T, the signal is at position x = - c T.
In stead of writing
x = - c t
you can just as well write
-x = c t, or
-12 x = 12 c t, or
exp(10 x) = exp(-10 c t)
Now you can safely forget everything else you ever
thought about this.
If want to become very stupid, you must continue
chatting with Oren Eleaticus Webster.
Dirk Vdm
eleaticus
> >> > ROFFLMFAO!
> >> > ROFFLMFAO!
> >> Tsk tsk.
Look, goofball, the subject matter in the immediate thread is the 'logic' of
including contradictories (that have a common solution only with t=0) in a
continuous derivation intended to apply to all values t>0.
Anything that does not address that concern/issue is a corrupt, evasive,
moronic irrelevancy.
If you want to assert your moronic derivations (true, I only looked at the
first, and it was super-moronic) then start a new thread, a thread that
isn't an evasion, a non-sequitor.
eleaticus
eleaticus
> So in the above equations, I would it understand better if x = ct
and -x
> = -ct. As it stands, I don't understand the meaning of the equation.
Einstein was trying to once again derive his x'=g(x-vt) and t'=g(t-vx/cc).
It is not possible to do so without (a) assuming the conclusion or (b)
mixing in absurdities.
Besides x=ct, x=-ct, and x=vt, three pairs of absurd contradictions for any
t<>0, he also asserted x'=ct' and x'=-ct' which are also mutual exclusive
for the intended t'<>0.
Also, of course, for the x' and t' he ends up with, it is not true that
x'=ct', nor x'=-ct'. But what's a couple more absurdities to the SR-cult?
He ends up with x'=g(x-vt).
Let v=.866c approx, so g=2. Let x=100 and t=100.
x'=g(x-vt) = 2(100 - .866(100))= 2(13.4)= 26.8.
x'=ct = 1(100)=100.
Just a wee contradiction, wouldn't you say?
But what is a mountain of absurdity to the likes of Uncle assAl, Dirk the
Dim, Ghost, etc?
My little tale about Dirk The Dim and his relentless search for the truth of
2+3=5 and the pencil law shows another example of the Eisteinian/Dirkian
absurdity.
> Thank you for your reply
You are welcome.
> Andre
eleaticus
>
>
Mich
Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
message news:ETlZd.37728$oC.35...@phobos.telenet-ops.be...
>
>
> Supposing c > 0,
> when you write x = c t , you describe a light signal
> going in the positive x direction, that happened to be
> at position x = 0 when your time read t = 0.
ok.
> The speed of the signal is c and the velocity is c as well.
> At time -T, the signal was at position x = - c T
> At time 0, the signal was at position x = 0
> At time T, the signal is at position x = c T.
ok, I see.
> In stead of writing
> x = c t
> you can just as well write
> -x = -c t, or
> -12 x = - 12 c t, or
> exp(10 x) = exp(10 c t)
ok.
>
> Still supposing c > 0,
> when you write x = - c t , you describe a light signal
> going in the negative x direction, that happened to be
> at position x = 0 when your time read t = 0.
ok
> The speed of the signal is c but the velocity is -c.
> At time -T, the signal was at position x = c T
> At time 0, the signal was at position x = 0
> At time T, the signal is at position x = - c T.
> In stead of writing
> x = - c t
> you can just as well write
> -x = c t, or
> -12 x = 12 c t, or
> exp(10 x) = exp(-10 c t)
ok.
> Now you can safely forget everything else you ever
> thought about this.
ok; thinking along this line does make sense, thanks.
> Dirk Vdm
Andre
Mich
> x is by definition not a scalar, it being a coordinate and hence must
change
> sign as the coordinate axis switches + and - directions.
>
> However, a scalar distance D would suffer even more, but more obviously.
>
> D=ct and D=-ct would make even SR-cult cretins sit up and think: "Oops!
That
> can't be for c>0 unless t=0!"
>
> eleaticus
thanks; I think I have it now.
Andre
Tom Capizzi
Once you attach directions to the variables, you are actually dealing with
vectors.
x = ct and x = -ct are not equivalent statements, but if you want to force
them to be,
then t in the second equation must be oppositely directed from t in the
first equation.
x = ct and -x = -ct contain the same information. The inclusion of minus
signs just
asserts that if you mirror reflect one side you must also mirror reflect the
other. The
variables x, c, and t still represent the same values in both equations. If
you want to
change the direction associated with a variable (by making it a vector),
then you
should rename the variables with subscripts or primes or whatever.
eleaticus
>
> "Mich" <mi...@efni.com> wrote in message
> news:1139ekk...@corp.supernews.com...
>
> > So in the above equations, I would it understand better if x = ct
> and -x
> > = -ct. As it stands, I don't understand the meaning of the equation.
>
> Einstein was trying to once again derive his x'=g(x-vt) and t'=g(t-vx/cc).
> It is not possible to do so without (a) assuming the conclusion or (b)
> mixing in absurdities.
>
> Besides x=ct, x=-ct, and x=vt, three pairs of absurd contradictions for
any
> t<>0, he also asserted x'=ct' and x'=-ct' which are also mutual exclusive
> for the intended t'<>0.
>
> Also, of course, for the x' and t' he ends up with, it is not true that
> x'=ct', nor x'=-ct'. But what's a couple more absurdities to the SR-cult?
>
> He ends up with x'=g(x-vt).
>
> Let v=.866c approx, so g=2. Let x=100 and t=100.
>
> x'=g(x-vt) = 2(100 - .866(100))= 2(13.4)= 26.8.
>
> x'=ct = 1(100)=100.
>
> Just a wee contradiction, wouldn't you say?
oops.
x'=ct'
t'=g(t-vx/cc)
t'=2(100-86.6)=26.8
x'=1(26.8)=26.8.
Nevermind!
eleaticus
David Cross
wrote:
>> He ends up with x'=g(x-vt).
>>
>> Let v=.866c approx, so g=2. Let x=100 and t=100.
>>
>> x'=g(x-vt) = 2(100 - .866(100))= 2(13.4)= 26.8.
>>
>> x'=ct = 1(100)=100.
>>
>> Just a wee contradiction, wouldn't you say?
>
>oops.
>
>x'=ct'
>
>t'=g(t-vx/cc)
>t'=2(100-86.6)=26.8
>
>x'=1(26.8)=26.8.
For someone who insisted that the primes on the times were irrelevant, you
certainly bothered to actually use them. Are you "corrupting" the
transformations now? ;)
---
David Cross
dcross1 AT shaw DOT ca
Tom Capizzi
>
> "Mich" <mi...@efni.com> wrote in message
> news:113b1rp...@corp.supernews.com...
>
>> > -x = -ct is the same as x=ct and thus provides no grounds for
>> > eventually
>> > concluding absurdly with the above transformations.
>
>> Ok...why are these two statements the same? In my mind, I thought x meant
> a
>> point at X distance from the observer in the direction - ( which could
> mean
>> North, for example), while + x would mean point, the same distance from
> the
>> observer, but in the opposite direction(South).
>> If that's the case, then, I don't see how both equations can mean the
>> same
>> thing.
>
> Hi, Mich!
>
> If -a = -b then a = b. Saying one says the other.
>
> It's the relationships among the terms that count, not the superficial
> differences that appear in the equations.
>
> a+c = b+c => a + b.
Typo: presumably a+c = b+c => a = b
>
> Neither of those two superficially different equation sets allows absurd
> algebraicly correct results.
>
> But x = -ct and x = ct does.
>
Quit making absurd claims. Nowhere in Einstein's 1905 paper does he make
the above claim. If you think he made it somewhere else, post a reference so
we can explain your error. Relativity does not depend on contradictions of
logic. It may assert apparent contradictions (so-called paradox), but these
vanish upon careful examination.
> For instance multiply side by side:
>
> x^2 = -(ct)^2
>
> x= ict.
>
> Where did i (imaginary root of -1) come from? It isn't implied or explicit
It came from your imagination, and has no bearing on the discussion.
> in either original equation; x,c,t are all real numbers. It is the result
> of
> the absurd contradictions inherent in the general simultaneous use of
> those
> equations. Yes, it too has a valid solution where t=0, but, again,
> Einstein
> was supposed to be deriving equations for all t>0, all x.
>
> As we have seen, just one equation, x=ct (x=-ct would also work) covers
> all
> values of x or t, for -c or +c.
The two equations describe two different light rays that start at the
origin. So
what? Would you feel better if they were x1 = ct and x2 = -ct?
eleaticus
"David Cross" <spamd...@nospam.net> wrote in message
news:h89c31hh6du7kdst1...@4ax.com...
Cross, you corruptoid, you, this thread is about the Lorentz-Einstein
transforms, the thread to which you allude is about the Newtonian-compatible
Galilean transforms.
If you'd actually think now and then instead of being so kind as to
regurgitate irrelevanices, you might have realized that.
eleaticus
David Cross
>> For someone who insisted that the primes on the times were irrelevant, you
>> certainly bothered to actually use them. Are you "corrupting" the
>> transformations now? ;)
>
> Cross, you corruptoid, you, this thread is about the Lorentz-Einstein
> transforms, the thread to which you allude is about the Newtonian-compatible
> Galilean transforms.
>
> If you'd actually think now and then instead of being so kind as to
> regurgitate irrelevanices, you might have realized that.
You have insisted that Maxwell's Equations obey the Galilean Transformations,
which rests on your explicit insistence that the primed times are
"irrelevancies". Shall I pull your old posts from google news for
verification?
So the fact that you now actually bothered to use the primed times and even
went back and fixed a mistake you made is, to me, more than a little amusing
given your holding-forth on the subject of the primed times.
I continue to find your use of strong language towards people who have not
used the same on you to be quite deplorable, by the way.
--
The Ghost In The Machine
<yDnZd.27556$6g7....@bignews1.bellsouth.net>:
Here, let me help you with that.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Now where, precisely, is the problem again?
Einstein eventually derives the equations transforming the
coordinate systems K (x,y,z,t) and k (xi, eta, zeta, tau)
to be
tau = beta * (t - vx/c^2)
xi = beta * (x - vt)
eta = y
zeta = z
where beta (most now use gamma) is 1/sqrt(1-v^2/c^2).
Einstein does make a restriction near the middle of section 3, which
leads to the constraint
x^2+y^2+z^2 = c^2t^2
or
xi^2 + eta^2 + zeta^2 = c^2tau^2
depending on coordinate system. This restriction doesn't appear
to cause much damage to the derivation.
There is a problem just after that point in introducing K', with
coordinates (x',y',z',t'). It turns out K' = K and thence
x' = x
y' = y
z' = z
t' = t.
Since Einstein defined x' = x-vt earlier, we have a scoping problem.
This may be it. Oh my!
Therefore, all of SR (and, since GR has SR as a special
case, GR as well) is thence invalidated and we'd better
all go back to Newton's theories, which, although totally
inconsistent with certain physical observations, don't
contain such awkward mathematical transitions.
Really. I mean, Einstein should have been more careful. He just
might have shown c' = c+v here.
</sarcasm>
The Ghost In The Machine
<spamd...@nospam.net>
wrote
on Mon, 14 Mar 2005 16:00:44 -0800
<h89c31hh6du7kdst1...@4ax.com>:
So did Einstein, it turns out. This gets *real* goofy fast, as
Einstein simultaneously defines x' = x-vt in one subpart of
Section 3, and then later finds that x' = x in a coordinate
system K' that turns out to be equal to K (k, the third
coordinate system, is the moving one here). This can of course
be explained away as a scoping problem, unless one is ultra-super-picky.
Like, erm, the previous poster in this subthread...
Of course nowadays most use (x,y,z,t) for the "stationary" coordinate
system, and (x',y',z',t') for the moving one. This isn't quite
according to Einstein, though in all fairness it's very hard to
render xi, eta, zeta, and tau in ASCII. :-)
All in all, the notation appears thorougly scrunged, and should probably
be carefully defined at the top of one's treatise... :-)
[.sigsnip]
Mich
Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
message news:ETlZd.37728$oC.35...@phobos.telenet-ops.be...
>
news:113ba73...@corp.supernews.com...
I was also wondering how one measures a moving frame.
Viewed through an analogy,I was thinking of measuring some distant
object/frame, called frame S.
The frame would be measured smaller than if it had been measured on the
frame itself, the difference being proportional to the distance.
If I claim the change in the measuring rods to be real, I then must also
assume a dilation of time, as the events on the frame also seem
slower.However,the time dilation is only due to our agreement in the real
change of measuring rods. For example, If a car on the far away frame S
travels from point A to point B at time t,according to the time measurement
observed on the frame S, it will be in agreement with my clock.Since,
however, I claim the car has travelled a shorter measured distance than what
an observer on frame S claimed was the distance traveled, I will conclude
his clock to be running slower than mine.Nevertheless, the time lapse of the
event remains the same.But, in the case of a time dilation and length
contraction caused by distance, we know it to be an illusory effect, and is
therefore not real.
So the next question; would the change in measurements caused by
velocity, real or simply illusory, as one measures a distant frame?
According to measurments of half life of pions, a relativistic dilation of
time seems to be detected.But is it a true dilation of time?What if the
velocity causes a moving frame to shrink in exactly in the same way distance
shrinks a frame? It would be only an illusionary effect, but, we would still
need to make the proper calculation, those found in relativity.In other
words, Could the frame of the moving pion (the distance it travelled) need
to be forshortened due to it's velocity relative the us(observer), in the
same way we would measure a far away frame to have been forshorten?
Has the time dilation observed on fast moving pions been confirmed by
clocks, or by
only the distance travelled?
Could distance and velocity have the same illusory effect on clocks and
measuring rods?
Andre
Lady Chatterly
Is that few of these things have anything to do with the cosmic
constants of this?
>Andre
He that plants thorns must never expect to gather roses.
--
Lady Chatterly
"So KK (he is almost smart enough to get his 3rd K) tell everyone what
you really think of Lady Chatterly..." -- Aratzio
Dirk Van de moortel
"Mich" <mi...@efni.com> wrote in message news:113f5ko...@corp.supernews.com...
>
> Dirk Van de moortel <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in
> message news:ETlZd.37728$oC.35...@phobos.telenet-ops.be...
> >
> > "Mich" <mi...@efni.com> wrote in message
> news:113ba73...@corp.supernews.com...
>
>
> I was also wondering how one measures a moving frame.
> Viewed through an analogy,I was thinking of measuring some distant
> object/frame, called frame S.
One does not measure frames.
One measures distances to and times of events.
If you send out a light signal at your clock time ts and receive
an echo at your clock time te, then you define the time of the
echo event according to you, as
t = 1/2 ( te + ts )
and the distance of the event according to you (at time t) as
x = 1/2 c ( te - ts ).
Note that this definition is equivalent with the conceptual method
of the frame with a grid of equaly spaced rigid measuring rods,
and synchronized clocks at every place.
Dirk Vdm
eleaticus
>
> "eleaticus" <elea...@bellsouth.net> wrote in message
> news:iAlZd.17664$c72....@bignews3.bellsouth.net...
> > If -a = -b then a = b. Saying one says the other.
> > It's the relationships among the terms that count, not the superficial
> > differences that appear in the equations.
> > a+c = b+c => a + b.
> Typo: presumably a+c = b+c => a = b
Yes, thank you.
> > Neither of those two superficially different equation sets allows absurd
> > algebraicly correct results.
> > But x = -ct and x = ct does.
> Quit making absurd claims.
Quite being an asshole.
> Nowhere in Einstein's 1905 paper does he make
> the above claim. If you think he made it somewhere else, post a reference
so
> we can explain your error.
This subthread started with Dirk The Dim's hacking of an article of mine,
which was about the 1916/1962 Appendix in which Albert made the screwups in
question.
(Let's see if you avoid the assholery a few of your fellow SR-cults enjoyed
evidencing.)
> Relativity does not depend on contradictions of
> logic.
Saying that is idiocy on your part.
x=ct plus x=-ct plus x=vt simultaneously in one continuous derivation
constitute three contradictions, not even counting the fact that their
common solution is for t=0, although the derivation is supposed to be for
all values t>=0.
>It may assert apparent contradictions (so-called paradox), but these
> vanish upon careful examination.
What you call careful examination are proofs by reduction to the absurd that
SR's coordinate transforms are ridiculous.
> > For instance multiply side by side:
> > x^2 = -(ct)^2
> > x= ict.
> > Where did i (imaginary root of -1) come from? It isn't implied or
explicit
> It came from your imagination, and has no bearing on the discussion.
No, idiot. It came directly from Einstein's assertion of both x=ct and
x=-ct.
> > in either original equation; x,c,t are all real numbers. It is the
result
> > of
> > the absurd contradictions inherent in the general simultaneous use of
> > those
> > equations. Yes, it too has a valid solution where t=0, but, again,
> > Einstein
> > was supposed to be deriving equations for all t>0, all x.
> > As we have seen, just one equation, x=ct (x=-ct would also work) covers
> > all
> > values of x or t, for -c or +c.
> The two equations describe two different light rays that start at the
> origin. So
> what?
So, if we were talking about a common solution, fine, but we are talking
about a derivation based on contradictions. A contradiction which leads
directly to the x=ict derivation above.
>Would you feel better if they were x1 = ct and x2 = -ct?
Of course, idiot, and then you couldn't misderive the transforms.
Tom Capizzi
>
> "Tom Capizzi" <etian...@verizon.net> wrote in message
> news:lLpZd.3120$qN3.1536@trndny01...
>>
>> "eleaticus" <elea...@bellsouth.net> wrote in message
>> news:iAlZd.17664$c72....@bignews3.bellsouth.net...
>
>> > If -a = -b then a = b. Saying one says the other.
>
>> > It's the relationships among the terms that count, not the superficial
>> > differences that appear in the equations.
>
>> > a+c = b+c => a + b.
>
>> Typo: presumably a+c = b+c => a = b
>
> Yes, thank you.
>
>> > Neither of those two superficially different equation sets allows
>> > absurd
>> > algebraicly correct results.
>
>> > But x = -ct and x = ct does.
>
>> Quit making absurd claims.
>
> Quite being an asshole.
>
Watch your language, sonny.
>> Nowhere in Einstein's 1905 paper does he make
>> the above claim. If you think he made it somewhere else, post a reference
> so
>> we can explain your error.
>
> This subthread started with Dirk The Dim's hacking of an article of mine,
> which was about the 1916/1962 Appendix in which Albert made the screwups
> in
> question.
>
I asked for a reference, not a political statement. I checked back in this
thread and
couldn't find a reference to the original Appendix. An Appendix which, by
the way, could
not have been written by Albert in 1962, since he was long dead.
> (Let's see if you avoid the assholery a few of your fellow SR-cults
> enjoyed
> evidencing.)
>
>> Relativity does not depend on contradictions of
>> logic.
>
> Saying that is idiocy on your part.
>
> x=ct plus x=-ct plus x=vt simultaneously in one continuous derivation
> constitute three contradictions, not even counting the fact that their
> common solution is for t=0, although the derivation is supposed to be for
> all values t>=0.
>
>>It may assert apparent contradictions (so-called paradox), but these
>> vanish upon careful examination.
>
> What you call careful examination are proofs by reduction to the absurd
> that
> SR's coordinate transforms are ridiculous.
>
>> > For instance multiply side by side:
>
>> > x^2 = -(ct)^2
>
>> > x= ict.
>
>> > Where did i (imaginary root of -1) come from? It isn't implied or
> explicit
>
>> It came from your imagination, and has no bearing on the discussion.
>
> No, idiot. It came directly from Einstein's assertion of both x=ct and
> x=-ct.
>
You contradict yourself here. You start with the premise that x = ct and x
= -ct.
Neither one of these is consistent with x = ict, since x, c and t are all
real. In
fact both equations x = ct and x = -ct are distinct solutions of x^2 =
c^2t^2.
What you have done through algebraic manipulations is to introduce a
spurious solution which came, as I said, from your imagination.
>> > in either original equation; x,c,t are all real numbers. It is the
> result
>> > of
>> > the absurd contradictions inherent in the general simultaneous use of
>> > those
>> > equations. Yes, it too has a valid solution where t=0, but, again,
>> > Einstein
>> > was supposed to be deriving equations for all t>0, all x.
>
>> > As we have seen, just one equation, x=ct (x=-ct would also work) covers
>> > all
>> > values of x or t, for -c or +c.
>
>> The two equations describe two different light rays that start at the
>> origin. So
>> what?
>
> So, if we were talking about a common solution, fine, but we are talking
> about a derivation based on contradictions. A contradiction which leads
> directly to the x=ict derivation above.
>
Only you are talking about contradictions. I'm sure that all you have done
is
to take equations out of context. Context which supplies the effective
subscripts
that I show below.
>>Would you feel better if they were x1 = ct and x2 = -ct?
>
> Of course, idiot, and then you couldn't misderive the transforms.
>
>> > No absurd result algebraicly possible.
>
> eleaticus
>
Post the pointer to the actual document so we can all see your claim vanish
or take your petty insults and foul language elsewhere. Your argument is
groundless.
eleaticus
>
> "eleaticus" <elea...@bellsouth.net> wrote in message
> news:lL_Zd.51282$%Y4.3...@bignews6.bellsouth.net...
> >> Quit making absurd claims.
> > Quite being an asshole.
> Watch your language, sonny.
Yep. 'Quite' is not correct, of course.
> >> Nowhere in Einstein's 1905 paper does he make
> >> the above claim. If you think he made it somewhere else, post a
reference
> > so
> >> we can explain your error.
> > This subthread started with Dirk The Dim's hacking of an article of
mine,
> > which was about the 1916/1962 Appendix in which Albert made the screwups
> > in
> > question.
> I asked for a reference, not a political statement. I checked back in this
> thread and
> couldn't find a reference to the original Appendix. An Appendix which, by
> the way, could
> not have been written by Albert in 1962, since he was long dead.
What part of "This subthread started with Dirk The Dim's hacking of an
article of mine" did you fail to understand?
> > (Let's see if you avoid the assholery a few of your fellow SR-cults
> > enjoyed
> > evidencing.)
The answer is in. You could not avoid the assholery of your fellow cult
cretins.
IE, what part of 1916/1962 did you not understand? Did the graf 1962
backward mask the 1916?
> > No, idiot. It came directly from Einstein's assertion of both x=ct and
> > x=-ct.
> You contradict yourself here. You start with the premise that x = ct and x
> = -ct.
> Neither one of these is consistent with x = ict, since x, c and t are all
> real.
Of course, you slimeball idiot. That was my point. My explicit point.
> In
> fact both equations x = ct and x = -ct are distinct solutions of x^2 =
> c^2t^2.
Asshole of an idiot. I gave no x^2 = c^2t^2. I just took Einstein's
simultaneous assertion of x=ct and x=-ct -- which you have been defending
(idiot!) -- and multiplied left side by left side, and right side by right
side, an obviously valid mathematical operation. That gives us:
x^2 = -(ct)^2.
And, x=ict.
The algebraic logic was soooo simple and absolutely impeccable, so the only
way we can get x=ict from x=ct and x=-ct is by the idiotic assertion of both
of the latter two simultaneously.
> What you have done through algebraic manipulations is to introduce a
> spurious solution which came, as I said, from your imagination.
Spurious solution? You are saying x=ict is not a solution of x^2=-(ct)^2?
What an idiot you are!
Spurious solution? You are finally admitting it was idiotic of Einstein to
simulaneously assert both x=ct and x=-ct? Those twin assertions lead
directly to your 'spurious' solution.
> >> The two equations describe two different light rays that start at the
> >> origin. So
> >> what?
God, you are an idiot!
> > So, if we were talking about a common solution, fine, but we are talking
> > about a derivation based on contradictions. A contradiction which leads
> > directly to the x=ict derivation above.
> Only you are talking about contradictions. I'm sure that all you have done
> is
> to take equations out of context. Context which supplies the effective
> subscripts
> that I show below.
ROFFLMFAO!
You cult cretins are like queued up lemmings. Right over the cliff, one
after another.
>
> >>Would you feel better if they were x1 = ct and x2 = -ct?
> >
> > Of course, idiot, and then you couldn't misderive the transforms.
> >
> >> > No absurd result algebraicly possible.
> >
> > eleaticus
> Post the pointer to the actual document so we can all see your claim
vanish
> or take your petty insults and foul language elsewhere. Your argument is
> groundless.
ROFFLMFAO!
You cult slimoids went through this same process just 2-3 months ago!
Look it up, idiot.
The difference was the latest (before you) cretins at least recognized that
Einstein's idiocy for what it was, delusionally thinking it was not
Einstein's but mine.
Look up 1962 + Einstein, idiot slimoid.
You have now posted maybe two dozen pieces in these threads and haven't
shown even an iota - or shadow thereof - of honesty in any of them. Not even
when totalled over all of them.
eleaticus
Tom Capizzi
>
> "Tom Capizzi" <etian...@verizon.net> wrote in message
> news:5o1_d.4508$ed6.2646@trndny06...
>>
>> "eleaticus" <elea...@bellsouth.net> wrote in message
>> news:lL_Zd.51282$%Y4.3...@bignews6.bellsouth.net...
>
[snip puerile rant]
> eleaticus
Your arguments lack any credibility. Nobody gives a rat's ass about
your opinion, you moron. You distort logic and pass it off as someone
else's error. You are the liar and a disrespectful, snot-nosed brat. And
that is my honest opinion. It is pointless to continue this discussion.
You are too stupid to understand how stupid you are. Big mouths
like you make a lot of noise, but all you really do is make a lot of wind.
You don't understand physics, you don't understand algebra, and you
have the arrogance to make derogatory claims about Einstein. You
can whine all you want, but you don't get to make the rules, and you
can't play by them, so goodbye.
eleaticus
> Your arguments lack any credibility. Nobody gives a rat's ass about
> your opinion, you moron. You distort logic and pass it off as someone
> else's error. You are the liar and a disrespectful, snot-nosed brat. And
> that is my honest opinion. It is pointless to continue this discussion.
> You are too stupid to understand how stupid you are. Big mouths
> like you make a lot of noise, but all you really do is make a lot of wind.
> You don't understand physics, you don't understand algebra, and you
> have the arrogance to make derogatory claims about Einstein. You
> can whine all you want, but you don't get to make the rules, and you
> can't play by them, so goodbye.
I gather that you finally got off your ass and found the Einstein material
you accused me of inventing.
Let's summarize.
I said that when Einstein (1916/1962) asserted x=ct, x=-ct, and x=vt
simultaneously in one continuous derviation process (to get his x'=g(x-vt),
etc) he was being absurd. (He also asserted, simultaneously, x'=ct' and
x'=-ct'.)
(A) You defended him, saying it was OK to do so.
(B) You continued to do so after reading my Dirk The Dim (Dimmy) fable which
gave an example of the kind of ludicrous result possible with such 'logic'.
(C) You told me I was using poor, spurious logic when I showed an example of
an impeccable continuous derviation from his assertions that even you agreed
led to a ridiculous result.
I multiplyed x=ct by x=-ct (left side by left side, etc) to get x^2=-(ct)^2
and x=ict. (That's multiplying equals by equals.)
Yet, even x=ct plus x=-ct is ludicrous for t<>0, which is the time range
goal of the derivation.
Your supposedly final idiocy is worth repeating in this context:
> Your arguments lack any credibility. Nobody gives a rat's ass about
> your opinion, you moron. You distort logic and pass it off as someone
> else's error. You are the liar and a disrespectful, snot-nosed brat. And
> that is my honest opinion. It is pointless to continue this discussion.
> You are too stupid to understand how stupid you are. Big mouths
> like you make a lot of noise, but all you really do is make a lot of wind.
> You don't understand physics, you don't understand algebra, and you
> have the arrogance to make derogatory claims about Einstein. You
> can whine all you want, but you don't get to make the rules, and you
> can't play by them, so goodbye.
I especially enjoy the rule that says you can assert mutually exclusive (for
t<>0) theora simultaneously in one continuous derivation.
I do hope you will publish your conclusion that the old standby, reduction
to the absurd, isn't valid because it is ok to not just reach
contradictions, but to include them as part of the logic.
eleaticus
Dirk Van de moortel
"eleaticus" <elea...@bellsouth.net> wrote in message news:nej_d.38501$5T6....@bignews4.bellsouth.net...
>
> "Tom Capizzi" <etian...@verizon.net> wrote in message
> news:AU8_d.6360$qN3.787@trndny01...
>
>
> > Your arguments lack any credibility. Nobody gives a rat's ass about
> > your opinion, you moron. You distort logic and pass it off as someone
> > else's error. You are the liar and a disrespectful, snot-nosed brat. And
> > that is my honest opinion. It is pointless to continue this discussion.
> > You are too stupid to understand how stupid you are. Big mouths
> > like you make a lot of noise, but all you really do is make a lot of wind.
> > You don't understand physics, you don't understand algebra, and you
> > have the arrogance to make derogatory claims about Einstein. You
> > can whine all you want, but you don't get to make the rules, and you
> > can't play by them, so goodbye.
>
> I gather that you finally got off your ass and found the Einstein material
> you accused me of inventing.
>
> Let's summarize.
>
> I said that when Einstein (1916/1962) asserted x=ct, x=-ct, and x=vt
> simultaneously in one continuous derviation process (to get his x'=g(x-vt),
> etc) he was being absurd. (He also asserted, simultaneously, x'=ct' and
> x'=-ct'.)
But you are an incredible idiot, aren't you?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html
and
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CorruptIdiocy.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorFiasco.html
and more.
DIrk Vdm
Tom Capizzi
>
> "Tom Capizzi" <etian...@verizon.net> wrote in message
> news:AU8_d.6360$qN3.787@trndny01...
>
>
>> Your arguments lack any credibility. Nobody gives a rat's ass about
>> your opinion, you moron. You distort logic and pass it off as someone
>> else's error. You are the liar and a disrespectful, snot-nosed brat. And
>> that is my honest opinion. It is pointless to continue this discussion.
>> You are too stupid to understand how stupid you are. Big mouths
>> like you make a lot of noise, but all you really do is make a lot of
>> wind.
>> You don't understand physics, you don't understand algebra, and you
>> have the arrogance to make derogatory claims about Einstein. You
>> can whine all you want, but you don't get to make the rules, and you
>> can't play by them, so goodbye.
>
> I gather that you finally got off your ass and found the Einstein material
> you accused me of inventing.
>
I did indeed attempt a search. I got a lot of hits but they were all written
about GR by others, not Einstein. Further, I never accused you of inventing
anything. I merely pointed out that Einstein was dead by 1962 and that you
don't understand the context in which the equations were written. If you
have a URL, post it.
> Let's summarize.
>
> I said that when Einstein (1916/1962) asserted x=ct, x=-ct, and x=vt
> simultaneously in one continuous derviation process (to get his
> x'=g(x-vt),
> etc) he was being absurd. (He also asserted, simultaneously, x'=ct' and
> x'=-ct'.)
>
In the first place, you have written a formula from Special Relativity,
1905.
In that paper, there is no such simultaneous assertion.
> (A) You defended him, saying it was OK to do so.
>
I said you misinterpreted what he had done, and took equations out of
context to create your own absurd derivation
.
> (B) You continued to do so after reading my Dirk The Dim (Dimmy) fable
> which
> gave an example of the kind of ludicrous result possible with such
> 'logic'.
>
Your attempt at 'humor' failed to amuse or interest me.
> (C) You told me I was using poor, spurious logic when I showed an example
> of
> an impeccable continuous derviation from his assertions that even you
> agreed
> led to a ridiculous result.
>
You don't know the meaning of impeccable. What you did is a classic grade
school algebra mistake.
> I multiplyed x=ct by x=-ct (left side by left side, etc) to get
> x^2=-(ct)^2
> and x=ict. (That's multiplying equals by equals.)
>
As you yourself pointed out, these variables only represent equals when
solved simultaneously, and then only in the trivial case when x = t = 0. In
that case even your ridiculous conclusion becomes true.
> Yet, even x=ct plus x=-ct is ludicrous for t<>0, which is the time range
> goal of the derivation.
>
The only useful observation here is that t can be negative. In any case, you
repeat your error of taking equations out of context.
> Your supposedly final idiocy is worth repeating in this context:
>
>> Your arguments lack any credibility. Nobody gives a rat's ass about
>> your opinion, you moron. You distort logic and pass it off as someone
>> else's error. You are the liar and a disrespectful, snot-nosed brat. And
>> that is my honest opinion. It is pointless to continue this discussion.
>> You are too stupid to understand how stupid you are. Big mouths
>> like you make a lot of noise, but all you really do is make a lot of
>> wind.
>> You don't understand physics, you don't understand algebra, and you
>> have the arrogance to make derogatory claims about Einstein. You
>> can whine all you want, but you don't get to make the rules, and you
>> can't play by them, so goodbye.
>
> I especially enjoy the rule that says you can assert mutually exclusive
> (for
> t<>0) theora simultaneously in one continuous derivation.
>
You don't get to invent your own rules, and you don't get to distort someone
else's argument by extracting pieces and claiming they are simultaneous.
> I do hope you will publish your conclusion that the old standby, reduction
> to the absurd, isn't valid because it is ok to not just reach
> contradictions, but to include them as part of the logic.
>
The only conclusion I can publish is that eleaticus is absurd.
> eleaticus
>
>
eleaticus
>
> "eleaticus" <elea...@bellsouth.net> wrote in message
> news:nej_d.38501$5T6....@bignews4.bellsouth.net...
> > I gather that you finally got off your ass and found the Einstein
material
> > you accused me of inventing.
> I did indeed attempt a search. I got a lot of hits but they were all
written
> about GR by others, not Einstein. Further, I never accused you of
inventing
> anything. I merely pointed out that Einstein was dead by 1962 and that you
> don't understand the context in which the equations were written.
The last part is pure lie.
And Einstein wasn't dead in 1916 (as in 1916/1962) so how is being dead in
1962 an excuse for your limemold behavior?
> > Let's summarize.
> >
> > I said that when Einstein (1916/1962) asserted x=ct, x=-ct, and x=vt
> > simultaneously in one continuous derviation process (to get his
> > x'=g(x-vt),
> > etc) he was being absurd. (He also asserted, simultaneously, x'=ct' and
> > x'=-ct'.)
> In the first place, you have written a formula from Special Relativity,
> 1905.
A formula from every paper on SR that ever existed. Funny that you think he
wouldn't use the formula in any paper but in the 1905 derivation.
> > (A) You defended him, saying it was OK to do so.
> I said you misinterpreted what he had done, and took equations out of
> context to create your own absurd derivation
That's a lie:
-------------
The two equations describe two different light rays that start at the
origin. So what? Would you feel better if they were x1 = ct and x2 = -ct?
-------------
.
> > (B) You continued to do so after reading my Dirk The Dim (Dimmy) fable
> > which
> > gave an example of the kind of ludicrous result possible with such
> > 'logic'.
> Your attempt at 'humor' failed to amuse or interest me.
My showing that simultaneous use of x=ct and x=-ct was absurd because it
would result in x=ict didn't amuse you either. The message was the same.
From x=ct and x=-ct we get x^2=-(ct)^2, and x=ict.
> > (C) You told me I was using poor, spurious logic when I showed an
example
> > of
> > an impeccable continuous derviation from his assertions that even you
> > agreed
> > led to a ridiculous result.
> You don't know the meaning of impeccable. What you did is a classic grade
> school algebra mistake.
The meaning is rather obvious when it applies to just two simple algebraic
operations:
(a) mutliplying of equals by equals, and
(b) taking the square roots of equals.
Since you consider those two operations as 'spurious', no wonder you have no
idea what impeccable means.
> > I multiplyed x=ct by x=-ct (left side by left side, etc) to get
> > x^2=-(ct)^2
> > and x=ict. (That's multiplying equals by equals.)
> As you yourself pointed out, these variables only represent equals when
> solved simultaneously, and then only in the trivial case when x = t = 0.
In
> that case even your ridiculous conclusion becomes true.
Not MY ridiculous conclusion. Einstein's. Implicit in Einstein's
simultaneous equations.
Einstein's derivation of his x' and t' using them is ridiculous except when
t=0.
> > Yet, even x=ct plus x=-ct is ludicrous for t<>0, which is the time range
> > goal of the derivation.
> The only useful observation here is that t can be negative. In any case,
you
> repeat your error of taking equations out of context.
ROFFLMFAO!
>>>. You
> >> can whine all you want, but you don't get to make the rules, and you
> >> can't play by them, so goodbye.
Gee. You deleted my 'supposed' final comment remark. How did I guess your
final wasn't your last?
> > I especially enjoy the rule that says you can assert mutually exclusive
> > (for
> > t<>0) theora simultaneously in one continuous derivation.
> You don't get to invent your own rules, and you don't get to distort
someone
> else's argument by extracting pieces and claiming they are simultaneous.
That's your rule. You said it was ok.
> > I do hope you will publish your conclusion that the old standby,
reduction
> > to the absurd, isn't valid because it is ok to not just reach
> > contradictions, but to include them as part of the logic.
> The only conclusion I can publish is that eleaticus is absurd.
LOL.
So, you admit that using x=ct and x=-ct simultaneously, with or without
x=vt, is ludicrous?
So, you admit that using x'=ct' and x'=-ct' simulataneously is ludicrous?
Yes or no, dipshit?
eleaticus
Tom Capizzi
In the 1905 paper there is no mention of x = ct and x = -ct.
>> > (A) You defended him, saying it was OK to do so.
>
>> I said you misinterpreted what he had done, and took equations out of
>> context to create your own absurd derivation
>
> That's a lie:
Prove it or shut up. Post a URL.
> -------------
> The two equations describe two different light rays that start at the
> origin. So what? Would you feel better if they were x1 = ct and x2 = -ct?
> -------------
> .
>> > (B) You continued to do so after reading my Dirk The Dim (Dimmy) fable
>> > which
>> > gave an example of the kind of ludicrous result possible with such
>> > 'logic'.
>
>> Your attempt at 'humor' failed to amuse or interest me.
>
> My showing that simultaneous use of x=ct and x=-ct was absurd because it
> would result in x=ict didn't amuse you either. The message was the same.
>
And just as irrelevant.
> From x=ct and x=-ct we get x^2=-(ct)^2, and x=ict.
>
>> > (C) You told me I was using poor, spurious logic when I showed an
> example
>> > of
>> > an impeccable continuous derviation from his assertions that even you
>> > agreed
>> > led to a ridiculous result.
>
>> You don't know the meaning of impeccable. What you did is a classic grade
>> school algebra mistake.
>
> The meaning is rather obvious when it applies to just two simple algebraic
> operations:
>
> (a) mutliplying of equals by equals, and
Wrong. They are only equals in the trivial case x = t = 0. Multiplication of
both sides of an equation by zero removes any valid information.
> (b) taking the square roots of equals.
>
Always introduces the possibility of extraneous solutions, because as
everyone
knows, there are two square roots for every positive number.
> Since you consider those two operations as 'spurious', no wonder you have
> no
> idea what impeccable means.
>
Speak for yourself.
>> > I multiplyed x=ct by x=-ct (left side by left side, etc) to get
>> > x^2=-(ct)^2
>> > and x=ict. (That's multiplying equals by equals.)
>
No. That's multiplication by zero.
>> As you yourself pointed out, these variables only represent equals when
>> solved simultaneously, and then only in the trivial case when x = t = 0.
> In
>> that case even your ridiculous conclusion becomes true.
>
> Not MY ridiculous conclusion. Einstein's. Implicit in Einstein's
> simultaneous equations.
>
Post the URL. You misrepresent Einstein. Don't expect us to fall for your
cheap trick.
> Einstein's derivation of his x' and t' using them is ridiculous except
> when
> t=0.
>
>> > Yet, even x=ct plus x=-ct is ludicrous for t<>0, which is the time
>> > range
>> > goal of the derivation.
>
>> The only useful observation here is that t can be negative. In any case,
> you
>> repeat your error of taking equations out of context.
>
> ROFFLMFAO!
>
You must not have much of an ass left, chuckles.
>>>>. You
>> >> can whine all you want, but you don't get to make the rules, and you
>> >> can't play by them, so goodbye.
>
> Gee. You deleted my 'supposed' final comment remark. How did I guess your
> final wasn't your last?
>
Keep up the personal attacks and continue getting replies. You have no
other arguments worth responding to.
>> > I especially enjoy the rule that says you can assert mutually exclusive
>> > (for
>> > t<>0) theora simultaneously in one continuous derivation.
>
>> You don't get to invent your own rules, and you don't get to distort
> someone
>> else's argument by extracting pieces and claiming they are simultaneous.
>
> That's your rule. You said it was ok.
>
Another lie.
>> > I do hope you will publish your conclusion that the old standby,
> reduction
>> > to the absurd, isn't valid because it is ok to not just reach
>> > contradictions, but to include them as part of the logic.
>
>> The only conclusion I can publish is that eleaticus is absurd.
>
> LOL.
>
> So, you admit that using x=ct and x=-ct simultaneously, with or without
> x=vt, is ludicrous?
>
> So, you admit that using x'=ct' and x'=-ct' simulataneously is ludicrous?
>
> Yes or no, dipshit?
>
Yes, you are a dipshit. Post the URL. What's the big deal? Chicken?
> eleaticus
>
>
eleaticus
> > From x=ct and x=-ct we get x^2=-(ct)^2, and x=ict.
> > The meaning is rather obvious when it applies to just two simple
algebraic
> > operations:
> > (a) mutliplying of equals by equals, and
> Wrong. They are only equals in the trivial case x = t = 0.
Moron. According to Einstein, x was equal to ct. I multiplied those two
equals by another pair of equals according to him: x and -ct, giving us x*x
= -ct*ct, x^2 = -(ct)^2, and x=ict.
> Multiplication of
> both sides of an equation by zero removes any valid information.
I agree, that is why I say Einstein committed multiple crimes against
science and logic.
> > b) taking the square roots of equals.
> Always introduces the possibility of extraneous solutions, because as
> everyone
> knows, there are two square roots for every positive number.
ROFFLMFAO!
You really think that -(ct)^2 is a positive number?
ROFFLMFAO!
eleaticus
Tom Capizzi
>
> "Tom Capizzi" <etian...@verizon.net> wrote in message
> news:UKq_d.3272$Ue6.1003@trndny04...
>
>> > From x=ct and x=-ct we get x^2=-(ct)^2, and x=ict.
>
>> > The meaning is rather obvious when it applies to just two simple
> algebraic
>> > operations:
>
>> > (a) mutliplying of equals by equals, and
>
>> Wrong. They are only equals in the trivial case x = t = 0.
>
> Moron. According to Einstein, x was equal to ct. I multiplied those two
> equals by another pair of equals according to him: x and -ct, giving us
> x*x
> = -ct*ct, x^2 = -(ct)^2, and x=ict.
>
So what? Einstein didn't multiply those two equations, you did. Any
absurdity
is yours.
>> Multiplication of
>> both sides of an equation by zero removes any valid information.
>
> I agree, that is why I say Einstein committed multiple crimes against
> science and logic.
>
Einstein didn't multiply both sides of an equation by zero, you did. The
absurdity is yours.
>> > b) taking the square roots of equals.
>
>> Always introduces the possibility of extraneous solutions, because as
>> everyone
>> knows, there are two square roots for every positive number.
>
> ROFFLMFAO!
>
> You really think that -(ct)^2 is a positive number?
It doesn't matter. There are two square roots for every negative number
as well. However, x^2 is a positive number. It doesn't matter anyway,
because the variables involved have only one value, zero, 0^2 = - 0^2,
and 0 = i 0.
>
> ROFFLMFAO!
>
> eleaticus
Post the URL, chicken.
Mich
me out a great deal, although I still do have certain problems.
eleaticus <elea...@bellsouth.net> wrote in message
news:iAlZd.17664$c72....@bignews3.bellsouth.net...
>
> "Mich" <mi...@efni.com> wrote in message
> news:113b1rp...@corp.supernews.com...
>
> > > -x = -ct is the same as x=ct and thus provides no grounds for
eventually
> > > concluding absurdly with the above transformations.
>
> > Ok...why are these two statements the same? In my mind, I thought x
meant
> a
> > point at X distance from the observer in the direction - ( which could
> mean
> > North, for example), while + x would mean point, the same distance from
> the
> > observer, but in the opposite direction(South).
> > If that's the case, then, I don't see how both equations can mean the
same
> > thing.
>
> Hi, Mich!
>
> If -a = -b then a = b. Saying one says the other.
Mathematically, I understand this, Eleaticus. Dirk started off very well by
giving me this explanation.
cut and paste:
-------------------------------------------------------------
"Supposing c > 0,
when you write x = c t , you describe a light signal
going in the positive x direction, that happened to be
at position x = 0 when your time read t = 0.
The speed of the signal is c and the velocity is c as well.
At time -T, the signal was at position x = - c T
At time 0, the signal was at position x = 0
At time T, the signal is at position x = c T."
----------------------------------------------------------------------
This part clarified the way as to how I ought to view the thought
experiement.I wasn't doing it right. Here, x is simply a unknown, so x = -ct
would mean x = to the coordinate -ct; if t is positive, then,
x will be -, that is the direction west from coordinate point x=0,for
example. This, I now can picture
he continues:
-------------------------------------
In stead of writing
x = c t
you can just as well write
-x = -c t,
-------------------------------
...so here I would assume -x is simply the coodinate point opposite
(coordinate point having the same distance from x=0 but opposite
direction )the coordinate point -x, which is indeed equal to x.
--------------------------------
>
> It's the relationships among the terms that count, not the superficial
> differences that appear in the equations.
>
> a+c = b+c => a + b.
>
> Neither of those two superficially different equation sets allows absurd
> algebraicly correct results.
ok.
>
> But x = -ct and x = ct does.
I agree, x in the first equation is not equal to x of the second equation.
It ought to be x1 and x2, it seems.
>
> For instance multiply side by side:
>
> x^2 = -(ct)^2
>
> x= ict.
>
> Where did i (imaginary root of -1) come from? It isn't implied or explicit
> in either original equation; x,c,t are all real numbers. It is the result
of
> the absurd contradictions inherent in the general simultaneous use of
those
> equations. Yes, it too has a valid solution where t=0, but, again,
Einstein
> was supposed to be deriving equations for all t>0, all x.
>
> As we have seen, just one equation, x=ct (x=-ct would also work) covers
all
> values of x or t, for -c or +c.
>
> No absurd result algebraicly possible.
I think I understand what you mean...but I have to say that my understanding
of physics is indeed very limited. I have problems visualizing simple
equations such as (x) *(-x); What does coordinate x
multiplied by coordinate -x actually mean? If I multiply (-x) * (y), the
result is an area...but (x) * (-x)
can only be a very long distance relative to the distance x.When can such a
equation be used?
> eleaticus
>
Andre