A simple analysis falsifies SR and LET

2 views
Skip to first unread message

MLuttgens

unread,
Nov 3, 1999, 3:00:00 AM11/3/99
to
Using LET, Wayne Throop has demonstrated (see ANNEX 1)
that the round-trip "ether" times (his "maxwellian" ether times)
along the arms of an interferometer are given by the formula

T(a) = (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2), where

L is the lenth of the arms,
v the velocity of the apparatus in the "ether", and
a the angle between the velocity vector and one of the arms,

and also that the corresponding interferometer times are

T(a)' = 2L/sqrt(1-v^2).

He stressed: "Relativity requires the arms to have no fringe
shift at any angle whatsoever, and the correct application of length
contraction arrives at exactly that conclusion".

According to him, length contraction by the Lorentz/SR factor
sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .
But a little later (see ANNEX 2), he made clear that "length contraction
didn't in fact affect the projection L*cos(a) of the arm, but instead
affected each strip across the arm in the direction of travel".

Indeed, length contraction of the projection of an arm by sqrt(1-v^2)
makes sense only if the arm itself is contracted by such factor.
Iow, due to its motion in the ether, the arm's length becomes
L*sqrt(1-v^2), and its projection on the x,x'-axis is of course
L*sqrt(1-v^2)*cos(a).

Alas for Wayne Throop, the arm's contraction CANNOT be independent from the
angle a that the arm makes with the velocity
vector.
If it were, an arm perpendicular to the velocity vector (a=90°) would still
be contracted by sqrt(1-v^2), in plain contradiction with SR or LET.

If the arms actually "have no fringe shift at any angle whatsoever",
Throop's formula T(a)' = 2L/sqrt(1-v^2) is correct, but then SR is falsified.

Interestingly enough, Paul B. Andersen, in his LET demonstration,
made the same mistake as Wayne Throop (see ANNEX 3).
For him also, the x-component of the rod is
Lx' =L*cos(a)*sqrt(1-v^2/c^2),
and the round trip time is 2L/sqrt(1 - v^2), hence independent of
the angle a.

And in his SR demonstration (see ANNEX 4), he derived the same
rond-trip time 2*L*g, where g= 1/sqrt(1-v^2), "confirming" that SR and
LET are experimentally indistinguishable (cf. Tom Roberts, ANNEX 5),
and BOTH wrong.

In fact, the round-trip time is not independent of the angle a, as
shown by my formula
T(a)' = (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2)
(see ANNEX 6).
Such formula is straightforwardly obtained if one assume that
the length contraction factor is sqrt (1-(v*cos(a))^2), thus not
independent of the angle a. In this factor, v*cos(a) is of course
the projection of the velocity vector v on the arm that makes an
angle a with the velocity vector.

Marcel Luttgens
------

ANNEX 1

From http://sheol.org/throopw/angle-light-bounce.gif

Write down the ether (or moving frame) coordinates of the
points light starts, is reflected, and stops.
Sum up the time out and the time back:

The x and y offsets are at right angles. Thus, distance is
sqrt(x^2+y^2). Without loss of generality, we have the light
pulse leaving the origin, and simply state
the coordinates of the other endpoint; that is, plug
(cos(a)*L*sqrt(1-v^2)+v*tO) for x, and (sin(a)*L) for y.
Divide by c=1, and you get the time out, tO. Same on the
reverse, with the obvious change in sign.

tO=sqrt((sin(a)*L)^2+(cos(a)*L*sqrt(1-v^2)+v*tO)^2) (1)
tB=sqrt((sin(a)*L)^2+(cos(a)*L*sqrt(1-v^2)-v*tB)^2) (2)
T(a) = tO+tB => 2L/sqrt(1-v^2)

Cos(a)*L is the x-extent of the arm. Length contraction acts
on the x-extents of objects. The application of length contraction
is exactly and precisely the division of x extents by gamma.

v*tO is the velocity times the time out,
v*tB is the velocity times the time back.

That's the relativistic or lorentzian case.
It's just the maxwellian case plus length contraction.

The corresponding "maxwellian" ether times are given by
T(a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2)

Both maxwellian and relativistic cases are derived from
calculating the distances between position of the arm
endpoints when light leaves an end, and the position of
the arm endpoints when light arrives at an end.

Relativity requires the arms to have no fringe shift at any
angle whatsoever, and the correct application of length
contraction arrives at exactly that conclusion.


ANNEX 2

From article <9318...@sheol.org>, Re: To Dennis, Keto and
all the Etherists, Mon, 12 Jul 1999 :
: How can length contraction affect the projection of a physical length,
: i.e. L*cos(a) ?

It doesn't. It affects each strip across the rod in the direction of
travel. Which ends up moving the x coordinate of the end of the rod
back to L*cos(a). And again, this is dead obvious. Consider the rod to
be a very long, thin rectangle, and now contract every infinitesimaly
small strip along some direction through the rectangle. Since the x
offset of each strip through the rectangle depends on the length of all
the strips before it, and those lengths are all contracted, the far end
of the rectangle is moved directly xward (with no change in y offset) to
be closer to the x offset of the near end. Of course, the resulting
shape is not a rectangle, but then, shapes aren't lorentz invariant;
that's also obvious.

ANNEX 3

In <380F8DE1...@hia.no>, Re: Limited applicability of SR,
Paul B. Andersen wrote:

I used the simplest way to arrive at the result:
the Lorentz transform. As I said, there are a number of ways
to attack the problem, but they all yield the same result
as long as you apply SR correctly.
If you insist on "using the contraction", here is how:
The y-component of the rod is: Ly' = L*sin(a)
The x-component of the rod is: Lx' = L*cos(a)*sqrt(1-v^2/c^2)
Let's find the "forth" light path:
Since the rod is moving, the end of the rod will move a distance
v*t1 in the x-direction while the light is on its way, where t1
is the time when the light hit.
Thus the x- and y-components of the light path will be:
Lp1x = Lx' + v*t1
Lp1y = Ly'
Since the light is moving with the speed c, we have:
(c*t1)^2 = (Lp1x)^2 + (Lp1y)^2
Inserting the equations above in this expression will give
a second order equation in t1, which can be solved.
I have done it, but I am to lazy to write down the calculations
in this awkward medium. But it is straight forward.
The result is: t1 = (L/c)*(1 + (v/c)*cos(a))/sqrt(1 - v^2/c^2)
thus Lp1 = L*(1 + (v/c)*cos(a))/sqrt(1 - v^2/c^2)

You can find the time back t2 in an equivalent way.
The only difference from the above is:
Lp2x = Lx' - v*t2
This will give the result:
t2 = (L/c)*(1 - (v/c)*cos(a))/sqrt(1 - v^2/c^2)
Lp2 = L*(1 - (v/c)*cos(a))/sqrt(1 - v^2/c^2)

So the round trip time is: t = (2L/c)/sqrt(1 - v^2/c^2),
independent of the the angle.

ANNEX 4

From article <37FBCA72...@hia.no>, Re: Limited
applicability of SR, Paul B. Andersen, Thu, 07 Oct 1999:

We have three events of interest:
E0: Light is emitted from end of the rod.
E1: Light is reflected off other end of the rod.
E2: Light returns to first end of the rod.

The co-ordinates of these events will in S' be:
E0: x0' = 0, y0' = 0, t0' = 0
E1: x1' = L*cos(a), y1' = L*sin(a), t1' = L/c
E2: x2' = 0, y2' = 0, t2' = 2*L/c

Now we can transform the co-ordinates of these events
to the S frame by the reverse LT:
x = g*(x' + v*t')
y = y'
t = g*(t' + x'*v/c^2)
where g = 1/sqrt(1-v^2/c^2)

We then get the co-ordinates:
(I only show the results, you can check the calculations
yourself, if you don't believe them.)

E0: x0 = 0, y0 = 0, t0 = 0
E1: x1 = g*L*(cos(a) + v/c), y1 = L*sin(a),
t1 = g*(L/c)*(1+(v/c)*cos(a))
E2: x2 = 2*g*L*(v/c), y2 = 0,
t2 = 2*g*L/c

The round trip time is of course t1-t0 = 2*g*L/c
It does not depend on the the angle a!
--------------------------------------

But let us see that everything checks out:

The length Lp1 of the light path forth is:
Lp1 = sqrt((x1-x0)^2 + (y1-y0)^2) = g*L*(1 + (v/c)*cos(a))
(Check it yourself if you don't believe it)
The length Lp2 of the light path back is:
Lp2 = srt((x2-x1)^2 + (y2-y1)^2) = g*L*(1 - (v/c)*cos(a))

Note that the length of the light pathes forth and back are
not equal. Quite obvious, if you think about it.

And of course, we have time forth tp1 = Lp1/c = t1
and time back tp2 = Lp2/c = t2 - t1

Note that these times are not equal!
Also quite obvious , if you think about it.

The total length of the light path Lp = Lp1 + Lp2 = g*2*L
It does not depend on a!

The round trip time is of course: Lp/c = g*2*L/c

ANNEX 5

From article <37CACC71...@lucent.com>, Re: Which contraction factor?,
Tom Roberts, Mon, 30 Aug 1999:

SR and LET are experimentally indistinguishable. _Provably_ so.
In fact, the situation is more dismal than that (as far as observing
the ether is concerned). Not only LET, but _every_ "reasonable" ether
theory (i.e. every member of the class of theories) is experimentally
indistinguishable from SR.

"reasonable" means "not already refuted by other experiments",
so the round-trip speed of light is isotropically c in every
intertial frame, for every theory belonging to this class.

ANNEX 6

The MMX, a general analysis (Version 5, October 1, 1999)
________________________

From a simple geometrical analysis of the light paths, it can
be shown that the round-trip "ether" time of light along the arms
of the interferometer is given by the formula

T(a) = (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2), where

L is the lenth of the arms,
v the velocity of the apparatus in the "ether", and
a the angle between the velocity vector and one of the arms.

Indeed, the sides of the right triangles that allows to
calculate tO (outgoing time of light along the arm) and tB
(time back), as well as their sum T(a) = tO+tB, are
(L*cos(a) + v*tO) or (L*cos(a) - v*tB), where L*cos(a)
is the projection of the arm on the x-axis, the height L*sin(a)
(the projection of the arm on the y-axis), and the hypotenuse
tO or tB.

Hence, solving the equations

tO=sqrt((sin(a)*L)^2+(cos(a)*L+v*tO)^2)
tB=sqrt((sin(a)*L)^2+(cos(a)*L-v*tB)^2),

one obtains

tO = (L/(1-v^2)) * sqrt(1-(v*sin(a))^2) + (L/(1-v^2)) * v*cos(a)
tB = (L/(1-v^2)) * sqrt(1-(v*sin(a))^2) - (L/(1-v^2)) * v*cos(a), and

T(a) = tO + tB = (2L/(1-v^2)) * sqrt(1-(v*sin(a))^2).

The arms of the MMX interferometer make an angle
of 90° between them, hence the "ether" times of the
perpendicular arm are given by

T(90+a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(90+a))^2, or

T(90+a) = (2*L/(1-v^2)) * sqrt(1-(v*cos(a))^2).

As Michelson and Morley observed no fringe shifts,
the difference between the times T(a)' and T(90+a)'
measured in the interferometer frame must be zero.

This is possible if

T(a)' = T(a) * f1 and T(90+a)' = T(90+a) * f2,
where f1=sqrt(1-(v*cos(a))^2) and f2=sqrt(1-(v*sin(a))^2).

Thus,
T(a)'= (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2)
T(90+a)' = (2*L/(1-v^2)) * sqrt(1-(v*cos(a))^2 * sqrt(1-(v*sin(a))^2),

and
T(a)' - T(90+a)' = 0.

Hence,the general formula giving the round-trip "interferometer" time
of light along the arms of the interferometer is

T(a)' = (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2).

That general formula, which is derived from purely logical
considerations, is confirmed by the following geometrical
analysis hypothesizing a length contraction of the arm:

Let's assume that L' = L * sqrt (1-(v*cos(a))^2) represents a length
contraction in the direction of the arm, because v*cos(a) is the
projection of the velocity vector v on the arm that makes an
angle a with the velocity vector.
Thus, the projection of the contracted arm on the x-axis is
L' *cos(a), and the sides of the right triangle that allows to
calculate tO' and tB', as well as their sum tO'+tB', are
(L' *cos(a) + v*tO) or (L' *cos(a) - v*tB), L' *sin(a), and
the hypotenuse tO' or tB'.

With the above length contraction, one obtains from

tO'=sqrt((L' *sin(a))^2+(L' *cos(a)+v*tO)^2), and
tB'=sqrt((L' *sin(a))^2+(L' *cos(a))-v*tB)^2)

the positive roots

tO' = tO * sqrt(1-(v*cos(a))^2)
tB' = tB * sqrt(1-(v*cos(a))^2), and of course

T(a)' = tO'+tB' =
(2L/(1-v^2)) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2),

which exactly matches the general formula obtained above
by simple logic.

This exact match justifies the above assumption that
the arm is contracted by sqrt(1-(v*cos(a))^2), and not by
sqrt(1-v^2).

Let's note that
T(a)' = (2L/(1-v^2)) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2),
can be expressed as
T(a)' = (2L/(1-v^2)) * sqrt( 1-v^2 + (v^4/4)*(sin(2a))^2),
a formula that is very close to T(a)' = 2L/sqrt(1-v^2),
which doesn't contain the angle a.

Consequences:

I. Perpendicular arms:

1) a=0°

Round-trip time of light along the parallel arm:

T(0)' = (2*L/(1-v^2) * sqrt(1-(v*sin(0))^2) * sqrt(1-(v*cos(0))^2)
= (2*L/(1-v^2) * 1 * sqrt(1-v^2)
= 2*L/sqrt(1-v^2)

Round-trip time of light along the perpendicular arm:

T(90)' = (2*L/(1-v^2) * sqrt(1-(v*sin(90))^2) * sqrt(1-(v*cos(90))^2)
= (2*L/(1-v^2) * sqrt(1-v^2) * 1
= 2*L/sqrt(1-v^2)

Thus, T(0)' = T(90)' = 2*L/sqrt(1-v^2)

2) a=30°

Round-trip time of light along the 30° arm:

T(30)' = (2*L/(1-v^2) * sqrt(1-(v*sin(30))^2) * sqrt(1-(v*cos(30))^2)
= (2*L/(1-v^2) * sqrt(1-(v*0.5)^2) * sqrt(1-(v*0.866)^2)

Round-trip time of light along the 120° arm:

T(120)' = (2*L/(1-v^2) * sqrt(1-(v*sin(120))^2) * sqrt(1-(v*cos(120))^2)
= (2*L/(1-v^2) * sqrt(1-(v*0.866)^2) * sqrt(1-(v*-0.5)^2)

As T(30)' = T(120)', there will be no fringe shift.

3) Let's rotate the apparatus to a=45°

Of course, this is the symmetrical case. For each arm,
the round-trip time will be T(45)', hence no fringe shift.

4) Let's again rotate the apparatus till a=90°

T(90)' = (2*L/(1-v^2) * sqrt(1-(v*sin(90))^2) * sqrt(1-(v*cos(90))^2)
= (2*L/(1-v^2) * sqrt(1-v^2) * 1
= 2*L/sqrt(1-v^2)

T(180)' = (2*L/(1-v^2) * sqrt(1-(v*sin(180))^2) * sqrt(1-(v*cos(180))^2)
= (2*L/(1-v^2)) * 1 * sqrt(1-v^2)
= 2*L/sqrt(1-v^2)

The times are again equal, thus no fring shift will be observed.

II. Arms making any angle between them.

If the other arm of the interferometer makes an angle alpha
with the arm having an angle a with the velocity vector,
we see immediately that
T(a+alpha)' = (2*L/(1-v^2) * sqrt(1-(v*sin(a+alpha))^2) *
sqrt(1-(v*cos(a+alpha))^2),
thus that the interferometer will lead to a null result
if alpha = 90° (case of the MMX).
Indeed, in that case, T(a)' = T(a+90)' for any value of a.

For any other value of alpha, T(a)' is generally different from
T(a+alpha), hence an interferometer whose arms make for
instance an angle of 45° between them should show fringe shifts
when rotated.

However, the anisotropy is so small that no present interferometer
is sensible enough to disclose it.

III. Detection of motion through the ether.

For "one-way" light trips and a=0° or 180°, one gets
from tB and tB' in the case of an interferometer of
equal opposite arms aligned along the velocity vector:

T(0°) = L/(1-v^2)) * (1-v) = L/(1+v)
T(180°) = L/(1-v^2)) * (1+v) = L/(1-v), and

T(0°)' = T(0°) * sqrt(1-v^2) = L * sqrt((1-v)/(1+v))
T(180°)' = T(180°) * sqrt(1-v^2) = L * sqrt((1+v)/(1-v)),

meaning that such "one-way" interferometer could detect
motion through the ether, even if time slowing is taken into
account.

The formula tB' = tB * sqrt(1-(v*cos(a))^2), with
tB = (L/(1-v^2)) * sqrt(1-(v*sin(a))^2) - (L/(1-v^2)) * v*cos(a),
is thus particularly interesting.

For small values of v (v<<c), tB' reduces to L*(1-v*cos(a)).
The time difference between opposite arms aligned along
the velocity vector is thus L*(1-v*cos(180)) - L*(1-v*cos(0)) =
L*(1+v) - L*(1-v) = 2L*v.

If the opposite arms are perpendicular to the velocity vector,
cos(90)=cos(270)=0, and the difference becomes zero.
For other inclinations of the opposite arms, the time differences
will be situated between 0 and 2L*v.

To the maximum time difference of 2L*v corresponds a phase
difference of 2L*v / lambda wavelengths.

Using a "one-way" interferometer with opposite arms of
L = 0.3 cm, a light of lambda = 6E-5 cm, and assuming
a velocity v = 1E-4 (thus 30 km/s) of the Earth through the ether,
a shift of N = 2*0.3*10^-4 / 6*10^-5 = 1 fringe is found for the
parallel case. Hence, by rotating the apparatus, the observed
shift should vary between 0 and 1 fringe.

An interesting experiment, using the "outward" formula
tO' = tO * sqrt(1-(v*cos(a))^2), where
tO = (L/(1-v^2)) * sqrt(1-(v*sin(a))^2) + (L/(1-v^2)) * v*cos(a),
has been proposed by Paul Stowe.

Its protocol is as follows:

We have two clocks (A & B) of sufficient accuracy as to
measure down to a precision of 0.1 nSec, and to match
any successive twenty-four hour rate to within 0.1 nSec.

Clock A is set to trigger a radio transmission on a x interval cycle
(frequency), continuously.

Clock B is linked to a receiver which records the precise
time that any of Clock A signals are received. That it, clock B
just say at HH:MM:SS.SSSSSSSSS I 'saw' a signal, over
& over again.

What one is looking for is, " any deviation in the reception
interval, period"!
Both clocks sit at fixed locations on the surface of the earth.

Marcel Luttgens

Wayne Throop

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
: mlut...@aol.com (MLuttgens)
: According to him, length contraction by the Lorentz/SR factor

: sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .

MLuttgens phrase "according to him" is misleading enough
to qualify as an intentional, malicious lie.

: But a little later (see ANNEX 2), he made clear that "length


: contraction didn't in fact affect the projection L*cos(a) of the arm,
: but instead affected each strip across the arm in the direction of
: travel".

Idiot. I doesn't DIRECTLY, PHYSICALLY affect the projection, since the
projection isn't physical. Instead, it PHYSICALLY affects the rod's
constituent parts (just as Lorentz described), and THAT physical
change (in LET analysis) results in a changed projection onto x.

No, not idiot. This is another case of being deliberately misleading.
Because MLuttgens was perfectly aware that the topic of discussion was
what length contraction did physically, and his objection was that
the projection of an arm isn't a physical thing, and so contraction
cannot act upon it.

So. Not an idiot at all: a deliberate liar.

: Alas for Wayne Throop, the arm's contraction CANNOT be independent


: from the angle a that the arm makes with the velocity vector.

I didn't say the ratio of L to L' was independent from the angle.
Nor does my derivation depend on "the arm's contraction" being
independent of the angle.

Which MLuttgens knows. And is lying about.

Doubtless, MLuttgens will post, and post, and post, and post,
and lie and weasel, and weasel and lie, until the universe
ends in heat-death. Feel free. I may periodically point out
the correct analysis that proves him wrong.

Just refer to the diagram

http://sheol.org/throopw/angle-light-bounce.gif

Wayne Throop thr...@sheol.org http://sheol.org/throopw

Rbwinn

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
>
>Idiot. I doesn't DIRECTLY, PHYSICALLY affect the projection, since the
>projection isn't physical. Instead, it PHYSICALLY affects the rod's
>constituent parts (just as Lorentz described), and THAT physical
>change (in LET analysis) results in a changed projection onto x.
>

Wayne,
What about my idiot? Remember the idiot who was at rest relative to a
star when it began emitting light? You were going to explain why a moving
observer was two different places when the star began emitting light, next to
the idiot observer and also at a point further away from the idiot observer.
Now , my idea was that the idiot observer was right, the moving observer
was at his position when the star began emitting light, the reason being that
his data agrees with data from the position of the star.
Now, you say that data in the frame of reference of the moving observer is
different, in that frame of reference, the star begins emitting light before
the moving observer reaches the position of the idiot observer.
The idiot observer and I think that the frame of reference of the star is
the best place to determine when the star begins emitting light.
Robert B. Winn

Wayne Throop

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
: rbw...@aol.com (Rbwinn)
: Remember the idiot who was at rest relative to a star when it began

: emitting light? You were going to explain why a moving observer was
: two different places when the star began emitting light, next to the
: idiot observer and also at a point further away from the idiot
: observer. Now , my idea was that the idiot observer was right, the
: moving observer was at his position when the star began emitting
: light, the reason being that his data agrees with data from the
: position of the star.

Let's be specific. Here's the scenario.

: rbw...@aol.com (Rbwinn) <19991011122543...@ng-bg1.aol.com>
: Now, the way I set up the problem was that the star emits light at t=0
: in the frame of reference of the star and at t=0 in the frame of
: reference of observer A. As far as what is happening in the frame of
: reference of observer B, so far we have no agreement whatsoever. We
: would need to take some time in observer B's frame of reference and
: call it t'=0. Now any time and place will do.

See http://sheol.org/throopw/rbwinn-diagram.gif
or http://sheol.org/throopw/rbwinn-diagram2.gif
for a simple diagram of the scenario being discussed.

Rbwinn (for some reason) doesn't understand that there is no PHYSICAL
EVIDENCE that is any more consistent with the idea that the star and A
are at rest, than with the idea that observer B is at rest.

And he hasn't yet pointed out any such PHYSICAL evidence. He's simply
said, in various ways "I think A is right, and you can't prove
otherwise". That's not physical evidence, that's just rbwinn stating a
prejudice. Because you can't prove B wrong, either.

Obvious, you can decide based on prejudice.
You just don't have any PHYSICAL EVIDENCE that is inconsistent
with B's coordinates, yet consistent with A's.

: The idiot observer and I think that the frame of reference of the star


: is the best place to determine when the star begins emitting light.

"best" for what reason? Other than your prejudice, I mean.

Of course, it's unclear what rbwinnis actually trying to say.
He refuses to learn a vocabulary precise enough to express the
controversy, preferring to muddle around in a fog of ambiguity.
What I've said is simple and straightforward. There is no physical
reason why A's rest coordinates are correct and B's are incorrect.
Rbwinn has never even tried to state what such a reason might be;
the closest he ever came to it was to "vote", because the rest
frames of the star and A are identical, so he could stuff the
ballot box. I hope nobody (else) is silly enough to think that
voting makes A's frame superior to B's?

Of course, A's distance is the *proper* distance.
But that couldn't be what rbwinn means, since he hasn't
a clue as to what a "proper distance" is.

Paul B. Andersen

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
MLuttgens wrote:

> According to him [Wayne Throop], length contraction by the Lorentz/SR factor


> sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .
> But a little later (see ANNEX 2), he made clear that "length contraction
> didn't in fact affect the projection L*cos(a) of the arm, but instead
> affected each strip across the arm in the direction of travel".

Resulting in that the x and y components of the length of moving
arm will be:
Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L



> Indeed, length contraction of the projection of an arm by sqrt(1-v^2)
> makes sense only if the arm itself is contracted by such factor.
> Iow, due to its motion in the ether, the arm's length becomes
> L*sqrt(1-v^2), and its projection on the x,x'-axis is of course
> L*sqrt(1-v^2)*cos(a).

So according to Marcel Luttgens, SR/LET predicts that the length of
a perpendicular arm (a = pi/2) would be L*sqrt(1-v^2).
It does not.



> Alas for Wayne Throop, the arm's contraction CANNOT be independent from the
> angle a that the arm makes with the velocity vector.

Alas? I would rather say fortunately since Wayne Throop say that the
length of the moving arm is:
L' = sqrt(Lx^2+Ly^2) = L*sqrt(1 - (v*cos(a))^2)
which is indeed dependent on the angle.

> If it were, an arm perpendicular to the velocity vector (a=90°) would still
> be contracted by sqrt(1-v^2), in plain contradiction with SR or LET.

Right. So Wayne Throop was right while you were wrong.



> If the arms actually "have no fringe shift at any angle whatsoever",
> Throop's formula T(a)' = 2L/sqrt(1-v^2) is correct, but then SR is falsified.

Uh? :-)
No fringe shifts falsify SR? :-)

> Interestingly enough, Paul B. Andersen, in his LET demonstration,
> made the same mistake as Wayne Throop (see ANNEX 3).
> For him also, the x-component of the rod is
> Lx' =L*cos(a)*sqrt(1-v^2/c^2),
> and the round trip time is 2L/sqrt(1 - v^2), hence independent of
> the angle a.

Right.
Since there is only one possible prediction from SR/LET and since
both Wayne and I are able to correctly calculate that prediction,
we obviously must get the same result.



> And in his SR demonstration (see ANNEX 4), he derived the same
> rond-trip time 2*L*g, where g= 1/sqrt(1-v^2), "confirming" that SR and
> LET are experimentally indistinguishable (cf. Tom Roberts, ANNEX 5),
> and BOTH wrong.

Sure SR and LET are indistinguishable.
That should surprice no one.



> In fact, the round-trip time is not independent of the angle a, as
> shown by my formula
> T(a)' = (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2)
> (see ANNEX 6).
> Such formula is straightforwardly obtained if one assume that
> the length contraction factor is sqrt (1-(v*cos(a))^2), thus not
> independent of the angle a. In this factor, v*cos(a) is of course
> the projection of the velocity vector v on the arm that makes an
> angle a with the velocity vector.

Oh, my dear. :-)
You keep going out of your way to find stumbling stones, Marcel.
One would expect this one was hard to find - but you did it!
And tripped.

So what have you done?
1. Correctly calculated the path length with no rod shortening,
e.g., a Galilean analysis.
This _path length_ is (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2),
which clearly and correctly depent on the angle.
Thats why Michelson expected of fringe shift.
2. Multiplied the _path length_ with the correct Lorents contraction
factor sqrt(1-(v*cos(a))^2) for the _arm_.

The error is:
It is not the light path length that is contracted by this factor,
------------------------------------------------------------------
it is the arm that is contracted.
---------------------------------

So you _oviously_ have to calculate the the path length _with_
the contracted arm.

Isn't the error glaringly obvious when it is pointed out to you,
Marcel? It should be. If not, I feel sorry for your.
Be honest to yourself, and you cannot fail to realize that I am
right. You have after all shown that you are able to calculate
the length of the light path when the length of the rod is given.
So all you have to do, is to realize that in a SR/LET analysis,
the given length of the rod is the contracted length.

Paul

Rbwinn

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
>And he hasn't yet pointed out any such PHYSICAL evidence. He's simply
>said, in various ways "I think A is right, and you can't prove
>otherwise". That's not physical evidence, that's just rbwinn stating a
>prejudice. Because you can't prove B wrong, either.
>

>
>Obvious, you can decide based on prejudice.
>You just don't have any PHYSICAL EVIDENCE that is inconsistent
>with B's coordinates, yet consistent with A's.

Wayne,
Well, I was hoping that it would not be necessary. In order for a star to
emit light, it has to be a quite massive object. For instance, the planet
Jupiter is a quite massive object, but not massive enough to emit light, so a
star is more massive than Jupiter.
Now, the idiot observer is not massive, neither does he emit light unless
he has a flashlight.
The moving observer, well, there is where we disagree. You are claiming
that the moving observer is at rest and the star is moving at velocity v, close
to the speed of light. That might be the reason why you have all these black
holes in your theory, since there would have to be some object more massive
than the star to cause the star to be moving at a rate of almost the speed of
light.
That is why I find it easier to have the star at rest than the moving
observer. Now this may seem prejudiced to you, but tell us exactly what it is
that is causing the star to move so fast relative to the moving observer.
Robert B.Winn

MLuttgens

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
In article <3821856E...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :


>Date : Thu, 04 Nov 1999 14:09:02 +0100


>
>MLuttgens wrote:
>
>> According to him [Wayne Throop], length contraction by the Lorentz/SR
>factor
>> sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .
>> But a little later (see ANNEX 2), he made clear that "length contraction
>> didn't in fact affect the projection L*cos(a) of the arm, but instead
>> affected each strip across the arm in the direction of travel".
>
>Resulting in that the x and y components of the length of moving
>arm will be:
>Lx = cos(a)*L*sqrt(1-v^2)
>Ly = sin(a)*L
>

Of course, who said otherwise?

>> Indeed, length contraction of the projection of an arm by sqrt(1-v^2)
>> makes sense only if the arm itself is contracted by such factor.
>> Iow, due to its motion in the ether, the arm's length becomes
>> L*sqrt(1-v^2), and its projection on the x,x'-axis is of course
>> L*sqrt(1-v^2)*cos(a).
>
>So according to Marcel Luttgens, SR/LET predicts that the length of
>a perpendicular arm (a = pi/2) would be L*sqrt(1-v^2).
>It does not.
>

Not according to Marcel Luttgens, but to simple logic.
If the arm is contracted by sqrt(1-v^2), thus independently
of the angle a, it will still be contracted by sqrt(1-v^2) if a=90°.
And this contradicts SR!

>> Alas for Wayne Throop, the arm's contraction CANNOT be independent from the
>> angle a that the arm makes with the velocity vector.
>
>Alas? I would rather say fortunately since Wayne Throop say that the
>length of the moving arm is:
>L' = sqrt(Lx^2+Ly^2) = L*sqrt(1 - (v*cos(a))^2)
>which is indeed dependent on the angle.
>

Where did he say that? For Wayne Throop, the length of the
moving arm is L*sqrt(1-v^2), and its x,x'-projection is
L*sqrt(1-v^2)*cos(a).
But I would be very happy if he had said, like I did, that its
contracted length is L*sqrt(1 - (v*cos(a))^2), and its projection is
L*sqrt(1 - (v*cos(a))^2)*cos(a).

>> If it were, an arm perpendicular to the velocity vector (a=90°) would still
>> be contracted by sqrt(1-v^2), in plain contradiction with SR or LET.
>
>Right. So Wayne Throop was right while you were wrong.
>

How can you claim the contrary of what has been said?
Are you so dishonest, or can't you understand what you read?



>> If the arms actually "have no fringe shift at any angle whatsoever",
>> Throop's formula T(a)' = 2L/sqrt(1-v^2) is correct, but then SR is
>falsified.
>
>Uh? :-)
>No fringe shifts falsify SR? :-)
>

I realize that you don't understand that Throop's formula (and yours)
is based on the false assumption that the moving arm is contracted
by sqrt(1-v^2), even when it is perpendicular to the velocity vector,
and hence is necessarily false.

>> Interestingly enough, Paul B. Andersen, in his LET demonstration,
>> made the same mistake as Wayne Throop (see ANNEX 3).
>> For him also, the x-component of the rod is
>> Lx' =L*cos(a)*sqrt(1-v^2/c^2),
>> and the round trip time is 2L/sqrt(1 - v^2), hence independent of
>> the angle a.
>
>Right.
>Since there is only one possible prediction from SR/LET and since
>both Wayne and I are able to correctly calculate that prediction,
>we obviously must get the same result.
>

So you agree here that you -and Wayne Throop- have used an arm
contracted by sqrt(1-v^2) in your demonstration. Then why did you
claim above that Throop has used L' = L*sqrt(1 - (v*cos(a))^2)?

You didn't read or understand, or didn't want to understand,
my general analysis!

I wrote:
Let's assume that L' = L * sqrt (1-(v*cos(a))^2) represents a length

contraction in the direction of the arm, because v*cos(a) is the


projection of the velocity vector v on the arm that makes an
angle a with the velocity vector.

Thus, the projection of the contracted arm on the x-axis is
L' *cos(a), and the sides of the right triangle that allows to
calculate tO' and tB', as well as their sum tO'+tB', are
(L' *cos(a) + v*tO) or (L' *cos(a) - v*tB), L' *sin(a), and
the hypotenuse tO' or tB'.

With the above length contraction, one obtains from

tO'=sqrt((L' *sin(a))^2+(L' *cos(a)+v*tO)^2), and
tB'=sqrt((L' *sin(a))^2+(L' *cos(a))-v*tB)^2)

the positive roots

tO' = tO * sqrt(1-(v*cos(a))^2)
tB' = tB * sqrt(1-(v*cos(a))^2), and of course

T(a)' = tO'+tB' =

(2L/(1-v^2)) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2).

Do you understand the sentence

"Thus, the projection of the contracted arm on the x-axis is
L' *cos(a), and the sides of the right triangle that allows to
calculate tO' and tB', as well as their sum tO'+tB', are
(L' *cos(a) + v*tO) or (L' *cos(a) - v*tB), L' *sin(a), and

the hypotenuse tO' or tB'." ?

If you did, you would realize that my derivation is similar
to yours or Throop's.
The only -and fundamental- difference is that I used
L * sqrt (1-(v*cos(a))^2) * cos(a) for the arm's projection,
instead of your false projection L * sqrt (1-v*^2) * cos(a).
False, because your projection necessarily implies an arm
contraction by sqrt(1-v^2) for the arm that is perpendicular
to the velocity vector.

Try to be honest! Don't obfuscate the issue!

>So you _obviously_ have to calculate the the path length _with_
>the contracted arm.
>

It is exactly what I did.

>Isn't the error glaringly obvious when it is pointed out to you,
>Marcel? It should be. If not, I feel sorry for your.
>Be honest to yourself, and you cannot fail to realize that I am
>right. You have after all shown that you are able to calculate
>the length of the light path when the length of the rod is given.
>So all you have to do, is to realize that in a SR/LET analysis,
>the given length of the rod is the contracted length.
>

Read correctly, solve the equations


tO'=sqrt((L' *sin(a))^2+(L' *cos(a)+v*tO)^2), and

tB'=sqrt((L' *sin(a))^2+(L' *cos(a))-v*tB)^2),
with L'= L * sqrt (1-(v*cos(a))^2),
and you will not repeat such gratuitous assertion.

>Paul

Marcel Luttgens

MLuttgens

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
In article <9416...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Thu, 04 Nov 1999 00:00:40 GMT
>
>: mlut...@aol.com (MLuttgens)
>: According to him, length contraction by the Lorentz/SR factor


>: sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .
>

>MLuttgens phrase "according to him" is misleading enough
>to qualify as an intentional, malicious lie.
>

>: But a little later (see ANNEX 2), he made clear that "length


>: contraction didn't in fact affect the projection L*cos(a) of the arm,
>: but instead affected each strip across the arm in the direction of
>: travel".
>

>Idiot. I doesn't DIRECTLY, PHYSICALLY affect the projection, since the
>projection isn't physical. Instead, it PHYSICALLY affects the rod's
>constituent parts (just as Lorentz described), and THAT physical
>change (in LET analysis) results in a changed projection onto x.
>

>No, not idiot. This is another case of being deliberately misleading.
>Because MLuttgens was perfectly aware that the topic of discussion was
>what length contraction did physically, and his objection was that
>the projection of an arm isn't a physical thing, and so contraction
>cannot act upon it.
>
>So. Not an idiot at all: a deliberate liar.
>

>: Alas for Wayne Throop, the arm's contraction CANNOT be independent


>: from the angle a that the arm makes with the velocity vector.
>

>I didn't say the ratio of L to L' was independent from the angle.
>Nor does my derivation depend on "the arm's contraction" being
>independent of the angle.
>
>Which MLuttgens knows. And is lying about.
>
>Doubtless, MLuttgens will post, and post, and post, and post,
>and lie and weasel, and weasel and lie, until the universe
>ends in heat-death. Feel free. I may periodically point out
>the correct analysis that proves him wrong.
>
>Just refer to the diagram
>
> http://sheol.org/throopw/angle-light-bounce.gif
>

But I refered to your diagram, and read
The Lorentzian or SR case (x extents foreshortened by gamma is
tO=sqrt((sin(a)*L)^2+(cos(a)*L*sqrt(1-v^2)+v*tO)^2)
etc...

And now, you are contradicting yourself:


"I didn't say the ratio of L to L' was independent from the angle.
Nor does my derivation depend on "the arm's contraction" being
independent of the angle."

Unless "x extents foreshortened by gamma" doesn't imply that
the arm of length L is contracted by gamma, and cos(a)*L*sqrt(1-v^2)
is not the x,x'-projection of L*sqrt(1-v^2)!
If this is what you are claiming now, I don't see how it is possible
to have a meaningful discussion with you.

>Wayne Throop

Marcel Luttgens

z@z

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
Hi Robert (B. Winn),

I think your simultaneity problem is quite interesting.

"We have a star that begins to emit light at t=0 in its own frame
of reference. We have an observer at rest, observer A, relative to
the star at distance d from the star. At t=0 in the frames of
reference of the star and observer A, two other observers reach
the position of observer A, observer B traveling toward the star at
a velocity of v and observer C traveling away from the star at a
velocity of v." http://www.deja.com/=dnc/getdoc.xp?AN=534639848

-->
observer C
-------o-----------------------:--------- t = 0
star observer A
< - - - - 10 LY - - - - >

Let us assume that the velocity of observer C (frame F') is

v = sqrt(0.9999)c = 0.99995 c --> gamma = 100

relative to the rest frame F. For observer A (frame F) the
situation is very simple. The star begins to emit light at t=0
which will be seen 10 years later.

The Lorentz transform with v = 0.99995 c gives:

[1] Dx' = 100 (Dx - 0.99995 Dt *c)
[2] Dt' = 100 (Dt - 0.99995 Dx /c)

[3] Dt = 100 (Dt' + 0.99995 Dx'*c)
[4] Dx = 100 (Dx' + 0.99995 Dt'/c)

Because in frame F simultaneity (i.e. Dt = 0) and a lenght of 10
LY (i.e. Dx = 10 LY) are given, we can calculate the corresponding
time difference in frame F' by using [2]:

Dt' = 100 (Dt - 0.99995 Dx ) = - 999.95 years

This means that a clock of observer C shows 999.95 years less
than a clock on the star when the star begins to emit light, if
both clocks are SR-synchronized in frame F'. Therefore for
observer C, the star will only in the far future (i.e. almost
1000 years later) begin to emit light.

For observer B however, which moves in the opposite direction
of C, the star has already long (999.95 years) ago started to
emit light.

The impossibility of such predictions becomes even more obvious
if we assume that observers B and C have only accelerated
shortly before t = 0.

Completely absurd are such predictions within (SR-style) LET.
What kind of ether properties could be responsible for the
creation of a time difference of almost 1000 years to an object
only 10 light years away, during a short acceleration phase?

Cheers, Wolfgang

Simultaneity and distance:
http://www.deja.com/=dnc/getdoc.xp?AN=543207365

Wayne Throop

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
:: you can decide based on prejudice. You just don't have any PHYSICAL

:: EVIDENCE that is inconsistent with B's coordinates, yet consistent
:: with A's.

: rbw...@aol.com (Rbwinn)
: Well, I was hoping that it would not be necessary.

You hoped you could get by on prejudice alone?

: In order for a star to emit light, it has to be a quite massive


: object. For instance, the planet Jupiter is a quite massive object,
: but not massive enough to emit light, so a star is more massive than
: Jupiter. Now, the idiot observer is not massive, neither does he emit
: light unless he has a flashlight.

And do you have a point?

: The moving observer, well, there is where we disagree.

Well DUH.

: You are claiming that the moving observer is at rest and the star is


: moving at velocity v, close to the speed of light.

I am claiming no such thing. Don't you even listen?

: That might be the reason why you have all these black holes in your


: theory, since there would have to be some object more massive than the
: star to cause the star to be moving at a rate of almost the speed of
: light.

We are talking SR, not GR. There are no black holes in SR.
Further, why do you need "something more massive"? For a star
to be moving "at lightspeed" requires nothing at all; even if you want
to discuss it in some center-of-momentum frame, a smaller mass could
be moving still faster in that frame.

But why even bother explaining? Rbwinn doesn't know what a frame is,
nor what coordinates are, nor anything about "black holes".

: That is why I find it easier to have the star at rest than the moving
: observer.

The universe simply doesn't care about rbwinn finds "easier".
Listing things rbwinn finds "easy" does not constitute any
physical evidence that the star is "at rest".

: Now this may seem prejudiced to you, but tell us exactly what it is


: that is causing the star to move so fast relative to the moving
: observer.

Inertia. Objects in motion tend to remain in motion. F=ma. All that.

Jonas

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
Rbwinn

You don't have to be sorry they will call anyone a moron who won't prefer
to solve problems using their context with relative frames which inherit
their invariant C(postulat). And in the "real world" there is many of us.

And by the way i really loved your sarcastic touch.
"So we have this little moving observer throwing a star around"
Love it :)

Wayne did not seem to like that at all ;)

<19991105084837...@ng-fe1.aol.com>...
>>
>>If you think I said ANYTHING REMOTELY LIKE that,
>>you're even stupider than I take you for.
>>
>
>Wayne,
> Well, you said I was a moron. I don't believe these inertial frames or
>whatever you call them. It may seem to work with only a star and an
observer,
>but suppose we take a universe of stars and rotate the observer relative to
the
>stars. Then the further away a star is, the faster it is moving relative
to
>the observer.
> Say, I think I have heard this idea before.
>Robert B. Winn

Jonas

unread,
Nov 4, 1999, 3:00:00 AM11/4/99
to
You never will beat these guys using their own math in their own examples
built by narrow views[local frames] relative to eachother. They will make
you come short out of the battle anytime you try.

You have to imagine/create another context where their approach are not
feasible. Instead of using the length of any realtime rod, use a sphere with
a radius and absolute time. Then it would be really hard for them to state
with a straight face that the length of the radius were contracted due to
the speed of the object, or that the real world sphere somehow was deformed.
I can't tell if the propagation of light is invariant C in Minowsky space
using the math of relativity, but it seems logically to me that it is.(This
due to the fact that they use their postulat invariant C in their math to
prove C is invariant).

In a real world object using Newtonian Spacetime(the sphere), this is not
the case however (so far as i can tell C is variant).


Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: mlut...@aol.com (MLuttgens)
: And now, you are contradicting yourself:

: "I didn't say the ratio of L to L' was independent from the angle.
: Nor does my derivation depend on "the arm's contraction" being
: independent of the angle."

See? My statements do not conflict in any way.
Where is this alleged self-contradiction?

: Unless "x extents foreshortened by gamma" doesn't imply that the arm


: of length L is contracted by gamma, and cos(a)*L*sqrt(1-v^2) is not
: the x,x'-projection of L*sqrt(1-v^2)!

"X extents forshortened by gamma" quite clearly does NOT imply that
"the arm of length L is contracted by gamma". As is obvious.

And cos(a)*L*sqrt(1-v^2) IS NOT the x,x'-projection of L*sqrt(1-v^2),
with "a" defined as I define it in the derivation.
As is also obvious.

: If this is what you are claiming now, I don't see how it is possible


: to have a meaningful discussion with you.

I'm not claiming it "now". I pointed it out from the very beginning.
The factor by which the length of the rod changes from measured to
coordinate value does not equal the factor by which the length of the
projection of that rod changes from measured to coordinate value,
because the angle of projection between x and x' also differs between
measured and coordinate values.

The issues MLuttgens brings up above were explained to him extensively
back when I thought he was honestly puzzled. The only way he can
still be claiming they are a problem is if he is, literally, a moron,
or if he's lying through his teeth.

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: mlut...@aol.com (MLuttgens)
: If the arm is contracted by sqrt(1-v^2), thus independently of the

: angle a, it will still be contracted by sqrt(1-v^2) if a=90 . And
: this contradicts SR!

Of course. That's why I never said the length of the arm was
independent of the angle. I was quite clear about this. It is the
light path length that is independent of the angle.

: For Wayne Throop, the length of the moving arm is L*sqrt(1-v^2),

Liar. I'm quite clear on this point. The length of the moving
arm is sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 ), where
a is the measured angle, not the coordinate angle.

sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 )
L*sqrt( sin(a)^2 + cos(a)^2 - (cos(a)*v)^2 )
L*sqrt( 1 - (cos(a)*v)^2 )

And I have never, ever, not once, said that the length of the moving
arm is L*sqrt(1-v^2) when at an angle to direction of motion, and
there's no way MLuttgens can honestly have mistaken what I've said.
He's intentionally lying about it.

: But I would be very happy if he had said, like I did, that its


: contracted length is L*sqrt(1 - (v*cos(a))^2), and its projection is
: L*sqrt(1 - (v*cos(a))^2)*cos(a).

Why should I make the same mistake MLuttgens makes?
The length is L*sqrt( 1 - (cos(a)*v)^2 ).
The projected length is L*cos(a)*sqrt(1-v^2).
The angle is not the coordinate angle, but this has been
extensively discussed, and somebody even worked the same problem
using coordinate angle instead of measured angle, and the two
methods were compared in some detail when MLuttgens claimed the
two analyses were in conflict. MLuttgens simply has no excuse
whatsoever for mistaking the simple facts of the matter.
He's intentionally lying about it.

He's a troll.

: So you agree here that you -and Wayne Throop- have used an arm


: contracted by sqrt(1-v^2) in your demonstration.

Of course not. Neither Paul nor I ever said the arm is contracted
by sqrt(1-v^2), and MLuttgens knows we never said that.
He's intentionally lying.

: Let's assume that L' = L * sqrt (1-(v*cos(a))^2) represents a length


: contraction in the direction of the arm,

OK. So we are not discussing relativity, nor lorentz ether theory.
Because, of co urse, the length L * sqrt (1-(v*cos(a))^2) does not
represent a contraction in the direction of the arm in either one
of those theories. As is obvious. As has been explained in
detail to MLuttgens. It is not plausible that he is still confused
on this simple point, so he must be lying. Trolling.

:: So you _obviously_ have to calculate the the path length
:: _with_ the contracted arm.
: It is exactly what I did. [...]
: you would realize that my derivation is similar to yours or Throop's.

: The only -and fundamental- difference is that I used L * sqrt
: (1-(v*cos(a))^2) * cos(a) for the arm's projection,

Paul obviously meant with the *correctly* contracted arm.
MLuttgens' contentions about the projected length of the arm
involve the counterfactual assumption that the measured angle
is equal to the coordinate angle.

: Try to be honest! Don't obfuscate the issue!

Follow your own advice. Quit lying. Quit trolling.

Rbwinn

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
>
>: Now this may seem prejudiced to you, but tell us exactly what it is
>: that is causing the star to move so fast relative to the moving
>: observer.
>
>Inertia. Objects in motion tend to remain in motion. F=ma. All that.
>

Wayne,
Thank you. Thank you very much. So we have this little moving observer
throwing a star around. But once again , the idiot observer says the star is
not moving, the moving observer is.
Now my question is, What if the moving observer was another idiot like the
observer at rest, and instead of saying, The star is moving., the way
scientists have trianed him, he says, I am moving. The star is standing still.
It seems to me that he could prove his statement because there would be some
cause for the motion, whether gravitation, a rocket engine, or whatever, which
could be measured to show that it was really the observer which was moving and
not the star.
Now my question is, Why can't the set of coordinates which represent the
moving observer be thought of as moving relative to the star from the position
of the moving observer?
Why does everyone have to have a frame of reference whichis stationary and
which causes all other things to move?
What would you call a set of cooridnates which is moving relative to a star
from its own position? In other words, if you say frame of reference,
scientists all get upset, since a frame of reference is not supposed to move,
but that is in itself very unrealistic because to properly represent a moving
observer, the set of coordinates which represent him has to move relative to
the star and the idiot observer.
Now you are trying to control the universe with your equations by getting
them to match up through distance contractions, etc. which I could care less
about. Einstein was talking about light, which I don't think he represented
correctly.
So my question is, when talking about light, why can't the moving observer
be represented by a moving set of coordinates
which are moving relative to a star instead of stationary coordinates which are
throwing a star somewhere?
Is it just me, or are there other people who think in terms of large
objects like stars, galaxies, etc. which move so little as to be considered
stationary compared to a moving observer moving relative to a star. I don't
really see the value of having all these large objects moving at velocities of
close to c relative to a moving observer. So I try to keep in mind what is
causing what to move. You call this prejudice, but I don't see how that
applies. It just seems to me that I am being realistic.
Robert B. Winn

Rbwinn

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
>Completely absurd are such predictions within (SR-style) LET.
>What kind of ether properties could be responsible for the
>creation of a time difference of almost 1000 years to an object
>only 10 light years away, during a short acceleration phase?
>
>Cheers, Wolfgang

Wolfgang,
Thank you for your response. I am only a welder, but the basic problem is
this: If you take a steel beam and put it on a truck and haul it to a jobsite,
is it shorter while it is being hauled than it is when it is sitting in the
steel yard?
The equations that scientists are using may be a shortcut to something, but
it is not to reality.
Let me ask you something: With regard to frames of reference, scientists
say that frames of reference always show a velocity of 0 for the thing they
represent. What do you call a set of coordinates which is moving relative to
something else when considered from its own position?
Robert B. Winn

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: rbw...@aol.com (Rbwinn)
: So we have this little moving observer throwing a star around.

If you think I said ANYTHING REMOTELY LIKE that,
you're even stupider than I take you for.

Wayne Throop thr...@sheol.org http://sheol.org/throopw

Paul B. Andersen

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
MLuttgens wrote:
>
> In article <3821856E...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Thu, 04 Nov 1999 14:09:02 +0100
> >
> >MLuttgens wrote:
> >
> >> According to him [Wayne Throop], length contraction by the Lorentz/SR
> >factor
> >> sqrt(1-v^2) acts on the x-extents of objects, i.e. on cos(a)*L .
> >> But a little later (see ANNEX 2), he made clear that "length contraction
> >> didn't in fact affect the projection L*cos(a) of the arm, but instead
> >> affected each strip across the arm in the direction of travel".
> >
> >Resulting in that the x and y components of the length of moving
> >arm will be:
> >Lx = cos(a)*L*sqrt(1-v^2)
> >Ly = sin(a)*L
> >
>
> Of course, who said otherwise?

Not Wayne. Not I. Remember that.



> >> Indeed, length contraction of the projection of an arm by sqrt(1-v^2)
> >> makes sense only if the arm itself is contracted by such factor.
> >> Iow, due to its motion in the ether, the arm's length becomes
> >> L*sqrt(1-v^2), and its projection on the x,x'-axis is of course
> >> L*sqrt(1-v^2)*cos(a).
> >
> >So according to Marcel Luttgens, SR/LET predicts that the length of
> >a perpendicular arm (a = pi/2) would be L*sqrt(1-v^2).
> >It does not.
> >
>
> Not according to Marcel Luttgens, but to simple logic.
> If the arm is contracted by sqrt(1-v^2), thus independently
> of the angle a, it will still be contracted by sqrt(1-v^2) if a=90°.
> And this contradicts SR!

Indeed it does contradict SR. And LET.
So this idea you say is wrong does not come from SR, LET, Wayne or me.
So _who_ does it come from then, if not you?


> >> Alas for Wayne Throop, the arm's contraction CANNOT be independent from the
> >> angle a that the arm makes with the velocity vector.
> >
> >Alas? I would rather say fortunately since Wayne Throop say that the
> >length of the moving arm is:
> >L' = sqrt(Lx^2+Ly^2) = L*sqrt(1 - (v*cos(a))^2)
> >which is indeed dependent on the angle.
> >
>
> Where did he say that?

Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L

"Who said otherwise?". Not Wayne.
Then L' = sqrt(Lx^2 + Ly^2) = L*sqrt(1-(v*cos(a))^2))

> For Wayne Throop, the length of the
> moving arm is L*sqrt(1-v^2), and its x,x'-projection is
> L*sqrt(1-v^2)*cos(a).

Ah!!! So that's where you got your wrong idea!
Lx = L*sqrt(1-v^2)*cos(a) is correct.
But this Lx is _not_ the x-projection of L*sqrt(1-v^2),
it is the x-projection of L*sqrt(1-(v*cos(a))^2))
You are mixing up the angles. A is _not_ the angle of the arm
in the stationary frame.
See below.

But honestly:
We say that the x-and y-projections are:

Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L

which clearly show that for a = pi/a, Lx = 0, Ly = L
And you claim that a = pi/2 yields the result:
Lx = 0, Ly = sqrt(1 - v^2)
How is it possible to miss that you must have screwed up
somewhere on your way to that conclusion? :-)

> But I would be very happy if he had said, like I did, that its
> contracted length is L*sqrt(1 - (v*cos(a))^2),

He did.

> and its projection is
> L*sqrt(1 - (v*cos(a))^2)*cos(a).

But this is wrong. You have yet again managed to dig up
a stumbling stone in which to trip.
You are a master in that respect! :-)



> >> If it were, an arm perpendicular to the velocity vector (a=90°) would still
> >> be contracted by sqrt(1-v^2), in plain contradiction with SR or LET.
> >
> >Right. So Wayne Throop was right while you were wrong.
> >
>
> How can you claim the contrary of what has been said?
> Are you so dishonest, or can't you understand what you read?

You puzzle me. Why are you saying these strange things?
It is obvious that Wayne Throop say above:
L' = L*sqrt(1 - (v*cos(a))^2)
("Who said otherwise?")
while you are arguing that "If the arm is contracted by sqrt(1-v^2),
thus independently of the angle a" is contrary to SR.
The latter is _your_ idea, nobody else have ever said this.
So then it is you that is wrong.
But you are right when you say that your idea is contrary to SR.



> >> If the arms actually "have no fringe shift at any angle whatsoever",
> >> Throop's formula T(a)' = 2L/sqrt(1-v^2) is correct, but then SR is
> >falsified.
> >
> >Uh? :-)
> >No fringe shifts falsify SR? :-)
> >
>
> I realize that you don't understand that Throop's formula (and yours)
> is based on the false assumption that the moving arm is contracted
> by sqrt(1-v^2), even when it is perpendicular to the velocity vector,
> and hence is necessarily false.

Indeed I don't understand that. Because it isn't.


Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L

"Who said otherwise?"
Then the arm is contracted by L' = L*sqrt(1-(v*cos(a))^2)

> >> Interestingly enough, Paul B. Andersen, in his LET demonstration,
> >> made the same mistake as Wayne Throop (see ANNEX 3).
> >> For him also, the x-component of the rod is
> >> Lx' =L*cos(a)*sqrt(1-v^2/c^2),
> >> and the round trip time is 2L/sqrt(1 - v^2), hence independent of
> >> the angle a.
> >
> >Right.
> >Since there is only one possible prediction from SR/LET and since
> >both Wayne and I are able to correctly calculate that prediction,
> >we obviously must get the same result.
> >
>
> So you agree here that you -and Wayne Throop- have used an arm
> contracted by sqrt(1-v^2) in your demonstration. Then why did you
> claim above that Throop has used L' = L*sqrt(1 - (v*cos(a))^2)?

This is getting ridiculous.


Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L

"Who said otherwise?"

Sigh.
It's getting a bit tiresome to show you all yourd weird errors.
Look again at what you say above.
We start with:


Lx = cos(a)*L*sqrt(1-v^2)
Ly = sin(a)*L

where Lx and Ly _are_ the x- and y-projections in the stationary frame,
and note well that a is the angle in the interferometer frame.
From this it is obvious that the contracted length is:
L' = sqrt(Lx^2 + Ly^2) = L*sqrt(1 - (v*cos(a))^2)

If you then take the x-projection of L' and get something which is
different from Lx, should it then not be blatantly obvious that you
must have made and error?
But not to Marcel Luttgens! He say:
Lx = L'*cos(a) which is different from Lx.
A beautiful self contradiction.

The error is that a is _not_ the angle of the rod in the stationary frame.
The angle of the rod a' in the stationary frame is different from a _beacuse_
the rod is contracted in the x-direction in this frame.
Thus:
Lx = L'*cos(a') = L*sqrt(1-v^2)*cos(a)
Ly = L'*sin(a') = L*sin(a)



> With the above length contraction, one obtains from
>
> tO'=sqrt((L' *sin(a))^2+(L' *cos(a)+v*tO)^2), and
> tB'=sqrt((L' *sin(a))^2+(L' *cos(a))-v*tB)^2)

Hey! What is tO' and tO here?
The time with and without contraction respectively?
You keep inventing new ways to screw up!
This is nonsense. You can obviously not mix two different
times in these equations.

The equations are:
tO = sqrt(Ly^2 + (Lx + v*tO)^2)
tB = sqrt(Ly^2 + (Lx - v*tB)^2)

You can solve them without contraction,
Lx = L*cos(a), Ly = L*sin(a)
in which case you get the result:
tO = (L/(1-v^2))*(sqrt(1-(v*sin(a))^2) + v*cos(a))
tB = (L/(1-v^2))*(sqrt(1-(v*sin(a))^2) - v*cos(a))

T(a) = tO + tB = (2L/(1-v^2))*sqrt(1-(v*sin(a))^2).

showing that the round trip time depend on the angle.
(The above are _your_ solutions in your ANNEX 6.
I have not checked them, but they seem correct.)

Or you can solve them with contraction,
Lx = L*cos(a)*sqrt(1-v^2), Ly = L*sin(a)
in which case you get the result:
tO = L*(1 + (v/c)*cos(a))/sqrt(1 - v^2)
tB = L*(1 - (v/c)*cos(a))/sqrt(1 - v^2)
T(a) = tO + tB = 2*L/sqrt(1 - v^2)
showing that the round trip do not depend on the angle.

> the positive roots
>
> tO' = tO * sqrt(1-(v*cos(a))^2)
> tB' = tB * sqrt(1-(v*cos(a))^2), and of course
>
> T(a)' = tO'+tB' =
> (2L/(1-v^2)) * sqrt(1-(v*sin(a))^2) * sqrt(1-(v*cos(a))^2).
>
> Do you understand the sentence
>
> "Thus, the projection of the contracted arm on the x-axis is
> L' *cos(a), and the sides of the right triangle that allows to
> calculate tO' and tB', as well as their sum tO'+tB', are
> (L' *cos(a) + v*tO) or (L' *cos(a) - v*tB), L' *sin(a), and
> the hypotenuse tO' or tB'." ?
>
> If you did, you would realize that my derivation is similar
> to yours or Throop's.
> The only -and fundamental- difference is that I used
> L * sqrt (1-(v*cos(a))^2) * cos(a) for the arm's projection,
> instead of your false projection L * sqrt (1-v*^2) * cos(a).
> False, because your projection necessarily implies an arm
> contraction by sqrt(1-v^2) for the arm that is perpendicular
> to the velocity vector.

But I suppose all your confusion should be cleared up now?



> Try to be honest! Don't obfuscate the issue!

:-)
Obfuscate - that is really the proper word for what you
are doing. The solution is if fact quite straight forward.
I am amazed of how many weird way you find to screw it up.



> >So you _obviously_ have to calculate the the path length _with_
> >the contracted arm.
> >
>
> It is exactly what I did.

If that was what you _thought_ you did, how come you got
the strange idea to use two different times (tO' and tO)
in your equations?
But anyway. Your "projections" were wrong.



> >Isn't the error glaringly obvious when it is pointed out to you,
> >Marcel? It should be. If not, I feel sorry for your.
> >Be honest to yourself, and you cannot fail to realize that I am
> >right. You have after all shown that you are able to calculate
> >the length of the light path when the length of the rod is given.
> >So all you have to do, is to realize that in a SR/LET analysis,
> >the given length of the rod is the contracted length.
> >
>
> Read correctly, solve the equations
> tO'=sqrt((L' *sin(a))^2+(L' *cos(a)+v*tO)^2), and
> tB'=sqrt((L' *sin(a))^2+(L' *cos(a))-v*tB)^2),
> with L'= L * sqrt (1-(v*cos(a))^2),
> and you will not repeat such gratuitous assertion.

But the question is of course, how do I solve four uknowns
from only two equations? :-)

I note however that the answer you gave:
T(a)' = (2*L/(1-v^2)*sqrt(1-(v*sin(a))^2)*sqrt(1-(v*cos(a))^2)
_is_ the path length we get with no contraction:
(2*L/(1-v^2)*sqrt(1-(v*sin(a))^2)
multiplied by the contaction factor of the arm sqrt(1-(v*cos(a))^2).
That is indisputable.

So how could I be wrong when I said that you:


1. Correctly calculated the path length with no rod shortening,
e.g., a Galilean analysis.
This _path length_ is (2*L/(1-v^2) * sqrt(1-(v*sin(a))^2),

2. Multiplied the _path length_ with the correct Lorents contraction
factor sqrt(1-(v*cos(a))^2) for the _arm_.

I do however now understand that you believed you did something else.
Exactly what, is still not clear to me.
But never mind. It's wrong anyway.

Paul

Rbwinn

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
>
>If you think I said ANYTHING REMOTELY LIKE that,
>you're even stupider than I take you for.
>

Wayne,


Well, you said I was a moron. I don't believe these inertial frames or
whatever you call them. It may seem to work with only a star and an observer,
but suppose we take a universe of stars and rotate the observer relative to the

stars. Then the further away a star is, the faster it is moving relative to

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: "Jonas" <la...@algonet.se>
: You don't have to be sorry they will call anyone a moron who won't

: prefer to solve problems using their context with relative frames

Nonsense. I could not possibly care less how you "solve your problems".
I only object when some self-important git tries to tell me how I should
be solving problems, or tries to tell me that SR is "inconsistent".

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: "Jonas" <la...@algonet.se>
: You never will beat these guys [...]

Why do you want to "beat these guys"?

: You have to imagine/create another context

Go ahead. I haven't noticed anybody holding you back.
But for some reason, you never seem to talk about your new
"context", all you can do is bad-mouth the classical approach
which has had a successful track record for more than 400 years.

: I can't tell if the propagation of light is invariant C in Minowsky
: space using the math of relativity,

Ah. Didn't take algebra in highschool?

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: "Paul B. Andersen" <paul.b....@hia.no>
: Ah!!! So that's where you got your wrong idea! Lx =

: L*sqrt(1-v^2)*cos(a) is correct. But this Lx is _not_ the
: x-projection of L*sqrt(1-v^2), it is the x-projection of
: L*sqrt(1-(v*cos(a))^2)) You are mixing up the angles. A is
: _not_ the angle of the arm in the stationary frame.

This has been pointed out to MLuttgens before.
He ignored it then, and I expect him to ignore it now.
Indeed, it is things like this, so simple that it is simply
implausible that he doesn't understand the issue once pointed out,
which convince me he's intentionally trolling, and getting his
jollies out of posing as an idiot and "fooling" other folks
into taking his pose seriously.

But that's just a conclusion based on his behavior, of course.

Wayne Throop

unread,
Nov 5, 1999, 3:00:00 AM11/5/99
to
: rbw...@aol.com (Rbwinn)
: I don't believe these inertial frames or whatever you call them.

Fine. I don't care what you believe, and I doubt the universe does either.

: It may seem to work with only a star and an observer, but suppose we


: take a universe of stars and rotate the observer relative to the
: stars.

Rotation is absolute, not relative. And of course analysis of an entire
universe full of stars and observers in terms of inertia and the con
cept of uniform motion being relative works very nicely, thank you very much.

MLuttgens

unread,
Nov 6, 1999, 3:00:00 AM11/6/99
to
In article <9417...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Fri, 05 Nov 1999 03:25:15 GMT


>
>: mlut...@aol.com (MLuttgens)
>: And now, you are contradicting yourself:
>: "I didn't say the ratio of L to L' was independent from the angle.
>: Nor does my derivation depend on "the arm's contraction" being
>: independent of the angle."
>
>See? My statements do not conflict in any way.
>Where is this alleged self-contradiction?
>
>: Unless "x extents foreshortened by gamma" doesn't imply that the arm
>: of length L is contracted by gamma, and cos(a)*L*sqrt(1-v^2) is not
>: the x,x'-projection of L*sqrt(1-v^2)!
>
>"X extents forshortened by gamma" quite clearly does NOT imply that
>"the arm of length L is contracted by gamma". As is obvious.
>
>And cos(a)*L*sqrt(1-v^2) IS NOT the x,x'-projection of L*sqrt(1-v^2),
>with "a" defined as I define it in the derivation.
>As is also obvious.
>
>: If this is what you are claiming now, I don't see how it is possible
>: to have a meaningful discussion with you.
>
>I'm not claiming it "now". I pointed it out from the very beginning.
>The factor by which the length of the rod changes from measured to
>coordinate value does not equal the factor by which the length of the
>projection of that rod changes from measured to coordinate value,
>because the angle of projection between x and x' also differs between
>measured and coordinate values.

>
>Wayne Throop

In article <9417...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Fri, 05 Nov 1999 03:56:57 GMT
>
>: mlut...@aol.com (MLuttgens)
>: If the arm is contracted by sqrt(1-v^2), thus independently of the


>: angle a, it will still be contracted by sqrt(1-v^2) if a=90 . And
>: this contradicts SR!
>

>Of course. That's why I never said the length of the arm was
>independent of the angle. I was quite clear about this. It is the
>light path length that is independent of the angle.
>
>: For Wayne Throop, the length of the moving arm is L*sqrt(1-v^2),
>
>Liar. I'm quite clear on this point. The length of the moving
>arm is sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 ), where
>a is the measured angle, not the coordinate angle.
>
> sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 )
> L*sqrt( sin(a)^2 + cos(a)^2 - (cos(a)*v)^2 )
> L*sqrt( 1 - (cos(a)*v)^2 )
>
>And I have never, ever, not once, said that the length of the moving
>arm is L*sqrt(1-v^2) when at an angle to direction of motion, and
>there's no way MLuttgens can honestly have mistaken what I've said.
>He's intentionally lying about it.
>

>: But I would be very happy if he had said, like I did, that its
>: contracted length is L*sqrt(1 - (v*cos(a))^2), and its projection is
>: L*sqrt(1 - (v*cos(a))^2)*cos(a).
>


>Why should I make the same mistake MLuttgens makes?
>The length is L*sqrt( 1 - (cos(a)*v)^2 ).
>The projected length is L*cos(a)*sqrt(1-v^2).
>The angle is not the coordinate angle, but this has been
>extensively discussed, and somebody even worked the same problem
>using coordinate angle instead of measured angle, and the two
>methods were compared in some detail when MLuttgens claimed the
>two analyses were in conflict. MLuttgens simply has no excuse
>whatsoever for mistaking the simple facts of the matter.
>He's intentionally lying about it.
>
>He's a troll.
>

>: So you agree here that you -and Wayne Throop- have used an arm


>: contracted by sqrt(1-v^2) in your demonstration.
>

>Of course not. Neither Paul nor I ever said the arm is contracted
>by sqrt(1-v^2), and MLuttgens knows we never said that.
>He's intentionally lying.
>

>: Let's assume that L' = L * sqrt (1-(v*cos(a))^2) represents a length


>: contraction in the direction of the arm,
>

>OK. So we are not discussing relativity, nor lorentz ether theory.

>Because, of course, the length L * sqrt (1-(v*cos(a))^2) does not


>represent a contraction in the direction of the arm in either one
>of those theories. As is obvious. As has been explained in
>detail to MLuttgens. It is not plausible that he is still confused
>on this simple point, so he must be lying. Trolling.

When a=90°or a=0°, the contracted length
L * sqrt (1-(v*cos(a))^2) is identical to that of SR/LET,
i.e. L*sqrt(1-v^2) or L.

>
>:: So you _obviously_ have to calculate the the path length
>:: _with_ the contracted arm.
>: It is exactly what I did. [...]
>: you would realize that my derivation is similar to yours or Throop's.

>: The only -and fundamental- difference is that I used L * sqrt
>: (1-(v*cos(a))^2) * cos(a) for the arm's projection,
>

>Paul obviously meant with the *correctly* contracted arm.
>MLuttgens' contentions about the projected length of the arm
>involve the counterfactual assumption that the measured angle
>is equal to the coordinate angle.
>

This is the key issue, and it should be solved mathematically,
not simply by words.
You said above:
"The length of the moving arm is
sqrt( (sin(a)*L)^2 +(sqrt(1-v^2)*cos(a)*L)^2 ), where


a is the measured angle, not the coordinate angle.
sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 )
L*sqrt( sin(a)^2 + cos(a)^2 - (cos(a)*v)^2 )
L*sqrt( 1 - (cos(a)*v)^2 )"

My question is, what is the formula giving the length of the
moving arm in terms of the coordinate angle?

>: Try to be honest! Don't obfuscate the issue!
>
>Follow your own advice. Quit lying. Quit trolling.
>

The origin of SR/LET problems is the silly idea that the
x,x'-projection of a moving arm is foreshortened by sqrt(1-v^2),
but not its y,y'-projection.
Think of an arm making an angle of 45° with the velocity vector V.
The situation is then perfectly symmetrical wrt the coordinate axes,
and the x,x'-projections must logically be identical to the
y,y'-projections.
But in SR/LET, they are different, i.e. Lx= L*sqrt(1-v^2)*sqrt(2)/2
and Ly=L*sqrt(2)/2 if one considers that V is parallel to x,x', or
Lx=L*sqrt(2)/2 and Ly= L*sqrt(1-v^2)*sqrt(2)/2 if V is perpendicular
to x,x'.This is of course nonsensical.
The correct approach is to consider that the arm is contracted by
sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
L' * sin(a).

>
>Wayne Throop

Marcel Luttgens

MLuttgens

unread,
Nov 6, 1999, 3:00:00 AM11/6/99
to
In article <3822DD25...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Fri, 05 Nov 1999 14:35:33 +0100


>
>MLuttgens wrote:
>>
>> In article <3821856E...@hia.no>, "Paul B. Andersen"
>> <paul.b....@hia.no> wrote :
>>
>> >Date : Thu, 04 Nov 1999 14:09:02 +0100
>> >

[I keep only the controversial point]

>> For Wayne Throop, the length of the
>> moving arm is L*sqrt(1-v^2), and its x,x'-projection is
>> L*sqrt(1-v^2)*cos(a).
>
>Ah!!! So that's where you got your wrong idea!
>Lx = L*sqrt(1-v^2)*cos(a) is correct.
>But this Lx is _not_ the x-projection of L*sqrt(1-v^2),
>it is the x-projection of L*sqrt(1-(v*cos(a))^2))
>You are mixing up the angles. A is _not_ the angle of the arm
>in the stationary frame.

>The angle of the rod a' in the stationary frame is different
>from a _because_the rod is contracted in the x-direction
> in this frame.

>We start with:


>Lx = cos(a)*L*sqrt(1-v^2)
>Ly = sin(a)*L
>where Lx and Ly _are_ the x- and y-projections in the
>stationary frame, and note well that a is the angle in the
>interferometer frame.
>From this it is obvious that the contracted length is:
>L' = sqrt(Lx^2 + Ly^2) = L*sqrt(1 - (v*cos(a))^2)
>

Please refer to my reply to Wayne Throop.

>Paul

Marcel Luttgens

MLuttgens

unread,
Nov 6, 1999, 3:00:00 AM11/6/99
to
In article <7vv4ij$qk0$1...@cubacola.tninet.se>, "Jonas" <la...@algonet.se> wrote
:


>Date : Thu, 4 Nov 1999 18:36:47 +0100


>
>You never will beat these guys using their own math in their own examples
>built by narrow views[local frames] relative to eachother. They will make
>you come short out of the battle anytime you try.

Thanks for your advice.
But if those guys were not SR-impaired, they would already
see that their theory is inconsistent.

Marcel Luttgens

Wayne Throop

unread,
Nov 7, 1999, 3:00:00 AM11/7/99
to
Refer to http://sheol.org/throopw/angle-light-bounce2.html
for a diagram of the situation, showing both a and a'.

:: MLuttgens' contentions about the projected length of the arm involve


:: the counterfactual assumption that the measured angle is equal to the
:: coordinate angle.

: mlut...@aol.com (MLuttgens)
: This is the key issue, and it should be solved mathematically,
: not simply by words.

OK. For the measured angle we have y/x=tan(a) by definition.
For the coordinate angle we have y'/x'=tan(a') for the same arm setup.
Since (per lorentz transform, or Lorentz length contraction) y=y', but
x=x'/sqrt(1-v^2), it is perfectly clear that a doesn't equal a'
(except for the obvious degenerate cases).

There. Mathematics.

: You said above:

: "The length of the moving arm is
: sqrt( (sin(a)*L)^2 +(sqrt(1-v^2)*cos(a)*L)^2 ), where
: a is the measured angle, not the coordinate angle.
: sqrt( (sin(a)*L)^2 + (sqrt(1-v^2)*cos(a)*L)^2 )
: L*sqrt( sin(a)^2 + cos(a)^2 - (cos(a)*v)^2 )
: L*sqrt( 1 - (cos(a)*v)^2 )"
: My question is, what is the formula giving the length of the
: moving arm in terms of the coordinate angle?:

Well, you can just substitute a'=atan(tan(a)/sqrt(1-v^2)),
then simplify. Cees Roos posted the correct form some time ago;
it's sqrt((1-v^2)/(1-(v*sin(a'))^2)

And, of course,

sqrt(1-(cos(a)*v)^2) = sqrt((1-v^2)/(1-(v*sin(a'))^2)

as I pointed out in <9341...@sheol.org>, in mid-august.

Wayne Throop

unread,
Nov 7, 1999, 3:00:00 AM11/7/99
to
: mlut...@aol.com (MLuttgens)
: The correct approach is to consider that the arm is contracted by

: sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and L'
: * sin(a).

No, only a slimy troll would claim that approach is "correct".
There is absolutely nothing in LET or SR that would indicate that
the angle of the rod is coordinate-invariant, and this has been
pointed out to MLuttgens many times before now.

He's not mistaken. He's lying. He's trying to mislead folks.

Paul B. Andersen

unread,
Nov 7, 1999, 3:00:00 AM11/7/99
to
MLuttgens wrote:
>
> The origin of SR/LET problems is the silly idea that the
> x,x'-projection of a moving arm is foreshortened by sqrt(1-v^2),
> but not its y,y'-projection.
> Think of an arm making an angle of 45° with the velocity vector V.
> The situation is then perfectly symmetrical wrt the coordinate axes,
> and the x,x'-projections must logically be identical to the
> y,y'-projections.

It is interesting to note that the consequence of your
demand that the x'- and y'- projections must be equal
when the x- and y-projections are equal, is that all projections
must change at the same ratio, and thus the contraction of the
arm cannot depend on it's orientation. The angle a does not matter.

And a moving circular object must according to you logically
be measured to be a smaller circular object.

Why is this "logically"?
What are the premises that leads to this remarkable conclusion?
Would you please state them?

> But in SR/LET, they are different, i.e. Lx= L*sqrt(1-v^2)*sqrt(2)/2
> and Ly=L*sqrt(2)/2 if one considers that V is parallel to x,x', or
> Lx=L*sqrt(2)/2 and Ly= L*sqrt(1-v^2)*sqrt(2)/2 if V is perpendicular
> to x,x'.

And a moving circular (in its rest frame) object will
be measured to be elliptical in the stationary frame.

> This is of course nonsensical.

Why is it nonsensical that the observed shortening is along
the direction of motion?

> The correct approach is to consider that the arm is contracted by
> sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
> L' * sin(a).

Why is "the correct approach" to assume that a moving circular object
is observed to be a smaller circular object in the stationary frame?
"Correct" according to what?
Are you referring to a specific theory which say this is "correct"?
Which theory is that?

Paul

Wayne Throop

unread,
Nov 7, 1999, 3:00:00 AM11/7/99
to
: "Paul B. Andersen" <paul.b....@hia.no>
: It is interesting to note that the consequence of [Mluttgens'] demand that
: the x'- and y'- projections must be equal when the x- and y-projections
: are equal, is that all projections must change at the same ratio, and
: thus the contraction of the arm cannot depend on it's orientation.
: The angle a does not matter.

Hm? I think you are wrong about that. He has said that the length
of the rod is L*(1-(v*cos(a))^2), and the angle invariant, which means
he gets by having the ratio y'/x' = y/x. But this doesn't imply
that the contraction factor cannot depend on orientation.

It means that the locus of locations of the far end of the rod
is not an elipse, as Lorentz requires, but it's an example that
fulfils equal projections, yet doesn't describe a circle.

But remember: it also isn't what Lorentz is talking about. Lorentz is
very clear, and states that physical objects are foreshortened along the
direction of translation, as shown in

http://sheol.org/throopw/angle-light-bounce2.html

which also shows why and how the angle is not invariant (that is, the
angle refered to the ether is not the angle as measured with comoving
instruments) in lorentz ether theory.

Paul B. Andersen

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
Wayne Throop wrote:
>
> : "Paul B. Andersen" <paul.b....@hia.no>
> : It is interesting to note that the consequence of [Mluttgens'] demand that
> : the x'- and y'- projections must be equal when the x- and y-projections
> : are equal, is that all projections must change at the same ratio, and
> : thus the contraction of the arm cannot depend on it's orientation.
> : The angle a does not matter.
>
> Hm? I think you are wrong about that. He has said that the length
> of the rod is L*(1-(v*cos(a))^2), and the angle invariant, which means
> he gets by having the ratio y'/x' = y/x. But this doesn't imply
> that the contraction factor cannot depend on orientation.

I think it does indeed imply that. (In a sensical world.)
The angle can only be invariant if the x- and y-projections
induvidually transform by multiplication by the same factor,
e.g. Lx' = f(v)*Lx, Ly' = f(v)*Ly. This means obviously that
the transformed length L' = f(v)*L.
M. Luttgens say that f(v) = L*(1-(v*cos(a))^2), e.g. depend
on the orientation of the line, which is self contradictory.

Think of the case where you have an arbitrary shaped object.
How do the lengths of the lines between arbitrary selected
points on this object transform? Think of three points arranged in
a triangle:
C
/|
/ |
/ |
A---B

It should be pretty obvious that when the angle of all three lines
shall be invariant, then the contraction factor cannot depend on the angle
of the lines. The lengths of all three lines must transform by multiplication
by the same factor f(v). So If the object is circular, it must
remain circular.
And it is quite nonsensical to say that the single contraction factor
according to which a circular object contracts depend on the angle of
some co-moving arm. :-)


> It means that the locus of locations of the far end of the rod
> is not an elipse, as Lorentz requires, but it's an example that
> fulfils equal projections, yet doesn't describe a circle.

Yes, but the implication is that the transform by which the shape
of objects transforms depend on the orientation of the _rod_.
If the rod is perpendicular to the motion, a circular object transforms
to a circular object of the same size, if the rod is parallel to the motion,
then the circular object transforms to a circular object contracted by 1/gamma.

The only consistent solution is that if the angle of the rod shall
be invariant, then the contraction factor cannot depend on the angle
of the rod.

Marcel Luttgens' claim is a huge self contradiction.
(As usual.)



> But remember: it also isn't what Lorentz is talking about. Lorentz is
> very clear, and states that physical objects are foreshortened along the
> direction of translation, as shown in
>
> http://sheol.org/throopw/angle-light-bounce2.html
>
> which also shows why and how the angle is not invariant (that is, the
> angle refered to the ether is not the angle as measured with comoving
> instruments) in lorentz ether theory.
>
> Wayne Throop thr...@sheol.org http://sheol.org/throopw

Paul

MLuttgens

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
In article <9419...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :


>Date : Sun, 07 Nov 1999 00:50:29 GMT

Thank you.

So, in SR/LET, the angular transformations are as follows:
sin (a') = sin(a) / sqrt(1-(v*cos(a)^2) (1)
cos(a') = cos(a) * sqrt(1-v^2) / sqrt(1-(v*cos(a)^2)) (2)
tan(a') = sin(a')/cos(a') = tan(a) / sqrt(1-v^2)

From relation (2), it can be calculated that the x,x'- projections
of the arm L*sqrt(1-(v*cos(a))^2)*cos(a') and L*sqrt(1-v^2)*cos(a)
are identical.
It should be clear that this is a direct consequence of the
ASSUMPTION that the y,y'-projection of the moving arm is
L*sin(a). Imo, this is only valid for the perpendicular arm.
In those expressions, a represents the angle between the arm
and the velocity vector, measured by an observer accompanying
the interferometer.

If the ether observer considers that the moving arm is
contracted by sqrt(1-(v*cos(a))^2), the perpendicular arm is
contracted by the factor 1, meaning that it is not contracted at all,
whereas the parallel arm is contracted by sqrt(1-v^2), in accordance
with LET and SR.
Instead of sqrt(1-(v*cos(a))^2), the ether observer could use
the contraction factor sqrt((1-v^2)/(1-(v*sin(a'))^2), but this
would change nothing, because relation (1) shows that those
two factors are strictly identical.
So, sqrt(1-(v*cos(a))^2) can be used by everybody as the arm
contraction factor.
However, using a or a' to calculate the projections of the contracted
arm leads to different calculated travel times of light along the arm,
but in both cases, the obtained results are compatible with the MMX.
Only new experiments using very sensible interferometers with
arms making between them an angle different from 90° could
settle the issue.

>
>Wayne Throop

Marcel Luttgens

MLuttgens

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
In article <9419...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :


>Date : Sun, 07 Nov 1999 03:10:54 GMT
>
>: mlut...@aol.com (MLuttgens)
>: The correct approach is to consider that the arm is contracted by
>: sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and L'
>: * sin(a).
>


>No, only a slimy troll would claim that approach is "correct".
>There is absolutely nothing in LET or SR that would indicate that
>the angle of the rod is coordinate-invariant, and this has been
>pointed out to MLuttgens many times before now.
>
>He's not mistaken. He's lying. He's trying to mislead folks.
>

Why did you snip this part?

The origin of SR/LET problems is the silly idea that the
x,x'-projection of a moving arm is foreshortened by sqrt(1-v^2),
but not its y,y'-projection.
Think of an arm making an angle of 45° with the velocity vector V.
The situation is then perfectly symmetrical wrt the coordinate axes,
and the x,x'-projections must logically be identical to the
y,y'-projections.

But in SR/LET, they are different, i.e. Lx= L*sqrt(1-v^2)*sqrt(2)/2
and Ly=L*sqrt(2)/2 if one considers that V is parallel to x,x', or
Lx=L*sqrt(2)/2 and Ly= L*sqrt(1-v^2)*sqrt(2)/2 if V is perpendicular

to x,x'.This is of course nonsensical.

The correct approach is to consider that the arm is contracted by
sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
L' * sin(a).

>
>Wayne Throop

Marcel Luttgens

MLuttgens

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
In article <38258056...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :


>Date : Sun, 07 Nov 1999 14:36:22 +0100


>
>MLuttgens wrote:
>>
>> The origin of SR/LET problems is the silly idea that the
>> x,x'-projection of a moving arm is foreshortened by sqrt(1-v^2),
>> but not its y,y'-projection.
>> Think of an arm making an angle of 45° with the velocity vector V.
>> The situation is then perfectly symmetrical wrt the coordinate axes,
>> and the x,x'-projections must logically be identical to the
>> y,y'-projections.
>

>It is interesting to note that the consequence of your


>demand that the x'- and y'- projections must be equal
>when the x- and y-projections are equal, is that all projections
>must change at the same ratio, and thus the contraction of the
>arm cannot depend on it's orientation. The angle a does not matter.
>

>And a moving circular object must according to you logically
>be measured to be a smaller circular object.
>
>Why is this "logically"?
>What are the premises that leads to this remarkable conclusion?
>Would you please state them?
>

Again you didn't read what I wrote!
I said: " Think of an arm making an angle of 45°...".

>> But in SR/LET, they are different, i.e. Lx= L*sqrt(1-v^2)*sqrt(2)/2
>> and Ly=L*sqrt(2)/2 if one considers that V is parallel to x,x', or
>> Lx=L*sqrt(2)/2 and Ly= L*sqrt(1-v^2)*sqrt(2)/2 if V is perpendicular

>> to x,x'.
>
>And a moving circular (in its rest frame) object will
>be measured to be elliptical in the stationary frame.
>

This is irrelevant, because I was speaking of an angle of 45°.

>> This is of course nonsensical.
>
>Why is it nonsensical that the observed shortening is along
>the direction of motion?
>

>> The correct approach is to consider that the arm is contracted by
>> sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
>> L' * sin(a).
>

>Why is "the correct approach" to assume that a moving circular object
>is observed to be a smaller circular object in the stationary frame?
>"Correct" according to what?
>Are you referring to a specific theory which say this is "correct"?
>Which theory is that?
>

Did you already forget that I just wrote


"The correct approach is to consider that the arm is contracted by
sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and

L' * sin(a)" ?
Hence, the moving object is not circular any more.

>Paul

Marcel Luttgens

MLuttgens

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
In article <3826D0F7...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Mon, 08 Nov 1999 14:32:39 +0100

Wayne Throop understands perfectly this issue, and he
is right. As usual, you have trouble in grasping what you read.

Marcel Luttgens

>
>Wayne Throop wrote:
>>
>> : "Paul B. Andersen" <paul.b....@hia.no>

>> : It is interesting to note that the consequence of [Mluttgens'] demand


>that
>> : the x'- and y'- projections must be equal when the x- and y-projections
>> : are equal, is that all projections must change at the same ratio, and
>> : thus the contraction of the arm cannot depend on it's orientation.
>> : The angle a does not matter.
>>

Wayne Throop

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
:: It means that the locus of locations of the far end of the rod is not

:: an elipse, as Lorentz requires, but it's an example that fulfils
:: equal projections, yet doesn't describe a circle.

: "Paul B. Andersen" <paul.b....@hia.no>
: Yes, but the implication is that the transform by which the shape of


: objects transforms depend on the orientation of the _rod_. If the rod
: is perpendicular to the motion, a circular object transforms to a
: circular object of the same size, if the rod is parallel to the
: motion, then the circular object transforms to a circular object
: contracted by 1/gamma.
:
: The only consistent solution is that if the angle of the rod shall be
: invariant, then the contraction factor cannot depend on the angle of
: the rod.

I still don't follow this. The *length* of the rod depends on the
*orientation* of the rod in both the case described by Lorentz, and that
described by MLuttgens. It's just that Lorentz keeps y=y', while
MLuttgens keeps a=a'. So why doesn't the above discussion mean that
LET-style length contraction cannot depend on the angle of the rod?

Mind you, this is a nit fairly far afield. It is still clear,
and I think we both agree, that whatever MLuttgens is discussing,
it isn't LET, or justified by anything in LET, nor (as he claims)
required by "logic".

Paul B. Andersen

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
MLuttgens wrote:
>
> In article <3826D0F7...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Mon, 08 Nov 1999 14:32:39 +0100
>
> Wayne Throop understands perfectly this issue, and he
> is right. As usual, you have trouble in grasping what you read.

You say that the transformed length of the rod should be
contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
shall remain a after the transformation.

There is nothing very special about a rod, so it must
be possible to transform other objects by the same transform.
So please explain how you think - say an quadratic object with
sides L in it's rest frame should transform.
(It could be a cube, but I am satisfied with a 2d analysis.)

L
A----------B
| |
L| |
| | -> v
| |
D----------C

What are the lengths of the sides and the diagonals in the frame
in which the frame is moving with the speed v?

Do _all_ the lengths contract by the same factor (1-(v*cos(a))^2)
dependent on some angle a? Which angle might a be? Hardly sensible.

So do _each_ distance contract by the factor (1-(v*cos(a))^2)
where a is the angle of that distance vector in such a way that the angle
of the distance vector remain invariant?

Please show me the shape of the resulting object.

Paul

Paul B. Andersen

unread,
Nov 8, 1999, 3:00:00 AM11/8/99
to
MLuttgens wrote:
>
> In article <38258056...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Sun, 07 Nov 1999 14:36:22 +0100
> >
> >MLuttgens wrote:
> >>
> >> The origin of SR/LET problems is the silly idea that the
> >> x,x'-projection of a moving arm is foreshortened by sqrt(1-v^2),
> >> but not its y,y'-projection.
> >> Think of an arm making an angle of 45° with the velocity vector V.
> >> The situation is then perfectly symmetrical wrt the coordinate axes,
> >> and the x,x'-projections must logically be identical to the
> >> y,y'-projections.
> >
> >It is interesting to note that the consequence of your

> >demand that the x'- and y'- projections must be equal
> >when the x- and y-projections are equal, is that all projections
> >must change at the same ratio, and thus the contraction of the
> >arm cannot depend on it's orientation. The angle a does not matter.
> >
> >And a moving circular object must according to you logically
> >be measured to be a smaller circular object.
> >
> >Why is this "logically"?
> >What are the premises that leads to this remarkable conclusion?
> >Would you please state them?
> >
>
> Again you didn't read what I wrote!
> I said: " Think of an arm making an angle of 45°...".

Sure you did.
And you demand that the angle shall remain 45 deg
after the transformation.
I suppose the demand is that any angle should remain invariant,
not only this particular one.
If I do not get your point wrong, it is that "all projections
must change at the same ratio", which _will_ leave the angle
invariant.



> >> But in SR/LET, they are different, i.e. Lx= L*sqrt(1-v^2)*sqrt(2)/2
> >> and Ly=L*sqrt(2)/2 if one considers that V is parallel to x,x', or
> >> Lx=L*sqrt(2)/2 and Ly= L*sqrt(1-v^2)*sqrt(2)/2 if V is perpendicular
> >> to x,x'.
> >
> >And a moving circular (in its rest frame) object will
> >be measured to be elliptical in the stationary frame.
> >
>
> This is irrelevant, because I was speaking of an angle of 45°.

Which is not invariant in LET/SR. That why the circle transforms
to an ellipse.



> >> This is of course nonsensical.
> >
> >Why is it nonsensical that the observed shortening is along
> >the direction of motion?
> >
> >> The correct approach is to consider that the arm is contracted by
> >> sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
> >> L' * sin(a).
> >
> >Why is "the correct approach" to assume that a moving circular object
> >is observed to be a smaller circular object in the stationary frame?
> >"Correct" according to what?
> >Are you referring to a specific theory which say this is "correct"?
> >Which theory is that?
> >
>
> Did you already forget that I just wrote
> "The correct approach is to consider that the arm is contracted by
> sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
> L' * sin(a)" ?
> Hence, the moving object is not circular any more.

If the circle is no circle anymore, then there must be distane vectors
in that circle which have changed their angle.

So it is self contradictory.

Se my other response to you in this thread.

Paul

Wayne Throop

unread,
Nov 9, 1999, 3:00:00 AM11/9/99
to
: mlut...@aol.com (MLuttgens)
: It should be clear that this is a direct consequence of the ASSUMPTION

: that the y,y'-projection of the moving arm is L*sin(a).

It's not an assumption. It's derived. In SR

[...] Thus phi(v)*phi(-v) = 1.
[...] Hence it follows that phi(v)=phi(-v).
--- Einstein, "electrodynamics of moving bodies" 1905, section 3

If you look at the context, you'll see phi(v) guarantees
that transforms of y-offsets remain unchanged.

Further, Lorentz arriveed at the same conclusion.

[...] so that we must put dl/dv=0; l=const.
The value of the constant must be unity, because we know
already tha for v=0, l=1.
--- Lorentz, "EM phenomena [in a moving system]" 1904, section 10

: Imo, this is only valid for the perpendicular arm.

Ok. Fine. So you aren't discussing either SR, nor LET.
So stop claiming you ARE discussing them, please.

Wayne Throop

unread,
Nov 9, 1999, 3:00:00 AM11/9/99
to
::: mlut...@aol.com (MLuttgens)
::: The correct approach is to consider that the arm is contracted by

::: sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
::: L' * sin(a).

:: thr...@sheol.org (Wayne Throop)
:: No, only a slimy troll would claim that approach is "correct". There
:: is absolutely nothing in LET or SR that would indicate that the angle
:: of the rod is coordinate-invariant, and this has been pointed out to


:: MLuttgens many times before now.
:: He's not mistaken. He's lying. He's trying to mislead folks.

: mlut...@aol.com (MLuttgens)
: Why did you snip this part?

( see <19991108125930...@ngol04.aol.com> )

Because the extra verbiage did not display anything in LET or SR
that would incicate that the angle is coordinate-invariant, so
MLuttgens' claim that keeping it invariant is "correct" in this
context is a deliberate lie, intended to mislead. Or rather, that's
what I conclude, since I normally operate under the assumption that
MLuttgens has more brain-power than the average sea-slug; with that
assumption, I'm forced to the conclusion that he lies intentionally.

Wayne Throop

unread,
Nov 9, 1999, 3:00:00 AM11/9/99
to
: "Paul B. Andersen" <paul.b....@hia.no>
: You [MLuttgens] say that the transformed length of the rod should be
: contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
: shall remain a after the transformation.

:
: There is nothing very special about a rod, so it must
: be possible to transform other objects by the same transform.
: So please explain how you think - say an quadratic object with

( "Quadratic object"? Hmmm. Let's see... quadratic equation,
quadratic function, quadratic surface, quadrilateral... nope.
I don't recall the term. I'm going to presume it means "square". )

: sides L in it's rest frame should transform.


: (It could be a cube, but I am satisfied with a 2d analysis.)
:
: L
: A----------B
: | |
: L| |
: | | -> v
: | |
: D----------C

Oooooh! A nice presentation.

Let me give a hint, OK? In particular, consider the contraction of
three rods, the first joining AD, the second joining DC, and the third
joining AC. Given that AD and DC contract by MLuttgens Rule, is it
logically possible for AC to do so? And if so, how?

Now try Lorentz' Rule. How well does that one work out?

Chortle.

Not that MLuttgens, being a troll, will admit the significance of this.
But I think any honest person can see it, once pointed out.


And this also means that, while a single rod in different positions
at different times can deform by MLuttgens' rule, and form a
self-consistent set locus of positions of the far end, and with the
contraction factor dependent on the angle, no objects with 2d extent can
do so. Which was probably what Paul meant, before; so we pretty much agree.

Paul B. Andersen

unread,
Nov 9, 1999, 3:00:00 AM11/9/99
to
Wayne Throop wrote:
>
> : "Paul B. Andersen" <paul.b....@hia.no>
> : You [MLuttgens] say that the transformed length of the rod should be
> : contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
> : shall remain a after the transformation.
> :
> : There is nothing very special about a rod, so it must
> : be possible to transform other objects by the same transform.
> : So please explain how you think - say an quadratic object with
>
> ( "Quadratic object"? Hmmm. Let's see... quadratic equation,
> quadratic function, quadratic surface, quadrilateral... nope.
> I don't recall the term. I'm going to presume it means "square". )

Right. Square.
But isn't "quadratic object" OK?
The first meaning of "quadratic" is in my Webster "square".
A square object would be an object formed as a square. :-)
But English is not my native language, and translating by dictonaries
does not always give a good result.

> : sides L in it's rest frame should transform.
> : (It could be a cube, but I am satisfied with a 2d analysis.)
> :
> : L
> : A----------B
> : | |
> : L| |
> : | | -> v
> : | |
> : D----------C
>
> Oooooh! A nice presentation.
>
> Let me give a hint, OK? In particular, consider the contraction of
> three rods, the first joining AD, the second joining DC, and the third
> joining AC. Given that AD and DC contract by MLuttgens Rule, is it
> logically possible for AC to do so? And if so, how?
>
> Now try Lorentz' Rule. How well does that one work out?
>
> Chortle.
>
> Not that MLuttgens, being a troll, will admit the significance of this.
> But I think any honest person can see it, once pointed out.
>
> And this also means that, while a single rod in different positions
> at different times can deform by MLuttgens' rule, and form a
> self-consistent set locus of positions of the far end, and with the
> contraction factor dependent on the angle, no objects with 2d extent can
> do so. Which was probably what Paul meant, before; so we pretty much agree.

Right.
One dimensional bodies do not exist in the real wold.
So it does not really work for a rod either.

Paul

Wayne Throop

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
:: "Quadratic object"? [] I'm going to presume it means "square".

: "Paul B. Andersen" <paul.b....@hia.no>
: Right. Square. But isn't "quadratic object" OK? The first meaning


: of "quadratic" is in my Webster "square". A square object would be an
: object formed as a square. :-) But English is not my native language,
: and translating by dictonaries does not always give a good result.

True. But refering to a dictionary anyways, my local electronic dictionary
(derived from the 1913 edition of websters IIRC, FWIW) says

quadratic
@qua.drat.ic \'kwa_:-'drat-ik\ adj : involving no higher power of
terms than a square <a quadratic equation>

The term "square" there isn't refering to shape, but a polynomial of
terms no more than second power. Similarly for other dictionaries; more
recent ones even say "terms of second degree" rather than "square".
Near as I can tell, "quadratic", when refering to shape, means "a shape
that can be expressed as an polynomial of power 2", which a square with
straight lines isn't one of.

But that doesn't mean "quadratic object" isn't perfectly sensible
usage; or that "square" wasn't a synonym for it in common usage somewhere
or sometime. It only means I haven't see it used that way before this.
And the main point is, I guessed right about the intended meaning,
so it can't be too bad a usage, eh?

MLuttgens

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
In article <9421...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :


>Date : Tue, 09 Nov 1999 01:52:52 GMT


>
>: "Paul B. Andersen" <paul.b....@hia.no>
>: You [MLuttgens] say that the transformed length of the rod should be
>: contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
>: shall remain a after the transformation.
>:
>: There is nothing very special about a rod, so it must
>: be possible to transform other objects by the same transform.
>: So please explain how you think - say an quadratic object with
>
> ( "Quadratic object"? Hmmm. Let's see... quadratic equation,
> quadratic function, quadratic surface, quadrilateral... nope.

> I don't recall the term. I'm going to presume it means "square". )


>
>: sides L in it's rest frame should transform.
>: (It could be a cube, but I am satisfied with a 2d analysis.)
>:
>: L
>: A----------B
>: | |
>: L| |
>: | | -> v
>: | |
>: D----------C
>
>Oooooh! A nice presentation.
>
>Let me give a hint, OK? In particular, consider the contraction of
>three rods, the first joining AD, the second joining DC, and the third
>joining AC. Given that AD and DC contract by MLuttgens Rule, is it
>logically possible for AC to do so? And if so, how?
>

I take a simple exemple:
Let L1=AD=1, and L2=DC=1
So, L3, the hypotenuse AC, is sqrt(2).
Let the velocity vector be parallel to L2, with v=0.8
Applying my contraction formula L'=L*sqrt(1-(v*cos(a))^2), we get:
For L1, a=90°, thus cos(a)=0, hence L1'=L1=1
For L2, a=0°, cos(a)=1, L2'=L2*sqrt(1-0.8^2)=0.6
For L3, a=180°-45°=135°, cos(a)= -sqrt(2)/2, (v*cos(a))^2=v^2/2,
and L3'=L3*sqrt(1-0.8^2/2)=sqrt(2)*sqrt(0.68)=1.1662
Checking:
We must have L3'^2 = L1'^2 + L2'^2, and indeed
1.1662^2 = 1^2 + 0.6^2 !

Who is the troll? Who are the trolls?

>Now try Lorentz' Rule. How well does that one work out?
>

You should try it yourself. You are the specialist of Lorenz!

>Chortle.

Ha Ha Ha !

>Not that MLuttgens, being a troll, will admit the significance of this.
>But I think any honest person can see it, once pointed out.
>
>
>And this also means that, while a single rod in different positions
>at different times can deform by MLuttgens' rule, and form a
>self-consistent set locus of positions of the far end, and with the
>contraction factor dependent on the angle, no objects with 2d extent can
>do so. Which was probably what Paul meant, before; so we pretty much agree.
>

See above.

>
>Wayne Throop

Marcel Luttgens

MLuttgens

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
In article <9421...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :


>Date : Tue, 09 Nov 1999 02:22:45 GMT

L'=L*sqrt(1-(v*cos(a))^2) is compatible with SR/LET.
More than compatible, it is a direct consequence of SR/LET,
as you have shown yourself.
But y=y' is only correct for perpendicular arms (see my other post).
A new SR/GR should be based on the contraction factor
sqrt(1-(v*cos(a))^2), and the constancy of the angle a.

>
>Wayne Throop

MLuttgens

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
In article <38274A2E...@hia.no>, "Paul B. Andersen"
<paul.b....@hia.no> wrote :

>Date : Mon, 08 Nov 1999 23:09:50 +0100
>
>MLuttgens wrote:
>>
>> In article <3826D0F7...@hia.no>, "Paul B. Andersen"
>> <paul.b....@hia.no> wrote :
>>

>> >Date : Mon, 08 Nov 1999 14:32:39 +0100
>>
>> Wayne Throop understands perfectly this issue, and he
>> is right. As usual, you have trouble in grasping what you read.
>

>You say that the transformed length of the rod should be


>contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
>shall remain a after the transformation.
>
>There is nothing very special about a rod, so it must
>be possible to transform other objects by the same transform.
>So please explain how you think - say an quadratic object with

>sides L in it's rest frame should transform.
>(It could be a cube, but I am satisfied with a 2d analysis.)
>
> L
> A----------B
> | |
> L| |
> | | -> v
> | |
> D----------C
>

>What are the lengths of the sides and the diagonals in the frame
>in which the frame is moving with the speed v?
>
>Do _all_ the lengths contract by the same factor (1-(v*cos(a))^2)
>dependent on some angle a? Which angle might a be? Hardly sensible.
>
>So do _each_ distance contract by the factor (1-(v*cos(a))^2)
>where a is the angle of that distance vector in such a way that the angle
>of the distance vector remain invariant?
>
>Please show me the shape of the resulting object.
>
>Paul

Please look at my reply to Wayne Throop.

Marcel Luttgens

MLuttgens

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
In article <9421...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :

>Date : Tue, 09 Nov 1999 02:47:15 GMT


>
>::: mlut...@aol.com (MLuttgens)
>::: The correct approach is to consider that the arm is contracted by
>::: sqrt(1-(v*cos(a))^2), and that its projections are L' * cos(a) and
>::: L' * sin(a).
>
>:: thr...@sheol.org (Wayne Throop)
>:: No, only a slimy troll would claim that approach is "correct". There
>:: is absolutely nothing in LET or SR that would indicate that the angle
>:: of the rod is coordinate-invariant, and this has been pointed out to
>:: MLuttgens many times before now.
>:: He's not mistaken. He's lying. He's trying to mislead folks.
>

In article <9420...@sheol.org>, thr...@sheol.org (Wayne Throop), you said:

"The *length* of the rod depends on the *orientation* of the rod in
both the case described by Lorentz, and that described by
MLuttgens. It's just that Lorentz keeps y=y', while MLuttgens
keeps a=a'."

This is indeed the only difference between LET/SR and me.
In both cases, the arm is contracted by sqrt(1-(v*cos(a))^2).

>: mlut...@aol.com (MLuttgens)
>: Why did you snip this part?
>
> ( see <19991108125930...@ngol04.aol.com> )
>
>Because the extra verbiage did not display anything in LET or SR
>that would incicate that the angle is coordinate-invariant, so
>MLuttgens' claim that keeping it invariant is "correct" in this
>context is a deliberate lie, intended to mislead. Or rather, that's
>what I conclude, since I normally operate under the assumption that
>MLuttgens has more brain-power than the average sea-slug; with that
>assumption, I'm forced to the conclusion that he lies intentionally.
>

I repeat, "Think of an arm making an angle of 45° with the velocity vector V.


The situation is then perfectly symmetrical wrt the
coordinate axes, and the x,x'-projections must logically be identical
to the y,y'-projections."

If there are preferred directions in space, y=y', thus a<>a'.
Otherwise, a=a', thus y<>y', and you are wrong.

>
>Wayne Throop

Marcel Luttgens

Paul B. Andersen

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
> >where a is the angle of that distance vector in such a way that the angle

> >of the distance vector remain invariant?
> >
> >Please show me the shape of the resulting object.
> >
> >Paul
>
> Please look at my reply to Wayne Throop.
>
> Marcel Luttgens

Done. As you will see of my response to that posting,
I am very pleased with your correct calculation of
how a square transforms according to Lorentz and SR.

I am looking forward to your reaction when you realize
what you have done. :-)

Paul

Paul B. Andersen

unread,
Nov 10, 1999, 3:00:00 AM11/10/99
to
MLuttgens wrote:
>
> In article <9421...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :
>
> >Date : Tue, 09 Nov 1999 01:52:52 GMT
> >
> >: "Paul B. Andersen" <paul.b....@hia.no>
> >: You [MLuttgens] say that the transformed length of the rod should be

> >: contracted by (1-(v*cos(a))^2), _and_ that the angle of the rod
> >: shall remain a after the transformation.
> >:
> >: There is nothing very special about a rod, so it must
> >: be possible to transform other objects by the same transform.
> >: So please explain how you think - say an quadratic object with
> >
> > ( "Quadratic object"? Hmmm. Let's see... quadratic equation,
> > quadratic function, quadratic surface, quadrilateral... nope.
> > I don't recall the term. I'm going to presume it means "square". )
> >
> >: sides L in it's rest frame should transform.

> >: (It could be a cube, but I am satisfied with a 2d analysis.)
> >:
> >: L
> >: A----------B
> >: | |
> >: L| |
> >: | | -> v
> >: | |
> >: D----------C
> >
> >Oooooh! A nice presentation.
> >
> >Let me give a hint, OK? In particular, consider the contraction of
> >three rods, the first joining AD, the second joining DC, and the third
> >joining AC. Given that AD and DC contract by MLuttgens Rule, is it
> >logically possible for AC to do so? And if so, how?
> >
>
> I take a simple exemple:
> Let L1=AD=1, and L2=DC=1
> So, L3, the hypotenuse AC, is sqrt(2).
> Let the velocity vector be parallel to L2, with v=0.8
> Applying my contraction formula L'=L*sqrt(1-(v*cos(a))^2), we get:
> For L1, a=90°, thus cos(a)=0, hence L1'=L1=1
> For L2, a=0°, cos(a)=1, L2'=L2*sqrt(1-0.8^2)=0.6
> For L3, a=180°-45°=135°, cos(a)= -sqrt(2)/2, (v*cos(a))^2=v^2/2,
> and L3'=L3*sqrt(1-0.8^2/2)=sqrt(2)*sqrt(0.68)=1.1662
> Checking:
> We must have L3'^2 = L1'^2 + L2'^2, and indeed
> 1.1662^2 = 1^2 + 0.6^2 !

Congratulations, you have just calculated the correct
contraction according to the Lorentz transform.



> Who is the troll? Who are the trolls?

Who indeed? :-)

Did you not notice that the transformed angle of L3' is:
a' = arccos(L1'/L3') = arccos(0.8575) = 149°
So the angles were hardly invariant, were they?

The square has contracted in the direction parallel to the motion,
but not in the direction transverse to motion, and then the angle
of the diagonal obviously has changed.

If we cut a strip out of the square along the diagonal, calling
it an "arm", then that strip would behave exactly like when the whole
square is present, don't you think?
(You obviously do not really have to cut away the rest of the square.
Who says we cannot use the diagonal of a square as the arm?)

So the angle of the arm is transformed, the parallel transformed
projection is shortened, the transverse is not.

Like Wayne and I have said all the time.



> >Now try Lorentz' Rule. How well does that one work out?
> >
>
> You should try it yourself. You are the specialist of Lorenz!

But you did it perfectly, Marcel!



> >Chortle.
>
> Ha Ha Ha !

Indeed. :-)



> >Not that MLuttgens, being a troll, will admit the significance of this.
> >But I think any honest person can see it, once pointed out.
> >
> >
> >And this also means that, while a single rod in different positions
> >at different times can deform by MLuttgens' rule, and form a
> >self-consistent set locus of positions of the far end, and with the
> >contraction factor dependent on the angle, no objects with 2d extent can
> >do so. Which was probably what Paul meant, before; so we pretty much agree.
> >
>
> See above.

Right.
Could not done it better myself.
Thanks.

Paul

Wayne Throop

unread,
Nov 11, 1999, 3:00:00 AM11/11/99
to
: mlut...@aol.com (MLuttgens)
: L'=L*sqrt(1-(v*cos(a))^2) is compatible with SR/LET.

But the presumption of an invariant angle is not.
As has been demonstrated beyond any rational dispute,
in other posts in this thread.

Wayne Throop

unread,
Nov 11, 1999, 3:00:00 AM11/11/99
to
: mlut...@aol.com (MLuttgens)
: I repeat, "Think of an arm making an angle of 45 with the velocity

: vector V. The situation is then perfectly symmetrical wrt the
: coordinate axes, and the x,x'-projections must logically be identical
: to the y,y'-projections." If there are preferred directions in space,
: y=y', thus a<>a'. Otherwise, a=a', thus y<>y', and you are wrong.

Of course, a<>a' doesn't imply a prefered direction in space; not even
in an ether theory; there is no prefered direction in the ether. It is
due to a direction of relative motion; the relative motion wrt the ether
in ether theories, of reference objects in SR. Which MLuttgens knows
very well. So again, he's lying. Dishonest through and through.

Further, let's look at how MLuttgens actually works a concrete problem.
Let's see if he actually keeps a=a'.

And
::: There is nothing very special about a rod, so it must


::: be possible to transform other objects by the same transform.
::: So please explain how you think - say an quadratic object with

::: sides L in it's rest frame should transform.


::: (It could be a cube, but I am satisfied with a 2d analysis.)
:::
::: L
::: A----------B
::: | |
::: L| |
::: | | -> v
::: | |
::: D----------C
::
::Oooooh! A nice presentation.
::
::Let me give a hint, OK? In particular, consider the contraction of
::three rods, the first joining AD, the second joining DC, and the third
::joining AC. Given that AD and DC contract by MLuttgens Rule, is it
::logically possible for AC to do so? And if so, how?

: I take a simple exemple:
: Let L1=AD=1, and L2=DC=1
: So, L3, the hypotenuse AC, is sqrt(2).
: Let the velocity vector be parallel to L2, with v=0.8
: Applying my contraction formula L'=L*sqrt(1-(v*cos(a))^2), we get:
: For L1, a=90°, thus cos(a)=0, hence L1'=L1=1
: For L2, a=0°, cos(a)=1, L2'=L2*sqrt(1-0.8^2)=0.6
: For L3, a=180°-45°=135°, cos(a)= -sqrt(2)/2, (v*cos(a))^2=v^2/2,
: and L3'=L3*sqrt(1-0.8^2/2)=sqrt(2)*sqrt(0.68)=1.1662
: Checking:
: We must have L3'^2 = L1'^2 + L2'^2, and indeed
: 1.1662^2 = 1^2 + 0.6^2 !

And what is the angle D'A'C' ?

In short, is a=a' as MLuttgens claims must occur?

No, in fact it is not. We have a'=atan(L1'/L2'), and that's
not 45 degrees as MLuttgens showed for his discussion of L3.
Because L1 and L2 are not equal.

So MLuttgens knows very well that he can't consistently do what
he claims must be done; he knows very well that the angle is not
invariant; he's just trying to mislead you. He's lying.

: Who is the troll? Who are the trolls?

MLuttgens' (mis)behavior speaks for itself.

# "Paul B. Andersen" <paul.b....@hia.no>
# I am looking forward to [MLuttgens'] reaction
# when [he] realize[s] what [he has] done. :-)

IMO he already knows what he has done.
And I expect he'll continue to lie and mislead to
try to keep other people from realizing it,
just as he has been doing all along.

Paul B. Andersen

unread,
Nov 11, 1999, 3:00:00 AM11/11/99