A simple analysis falsifies SR and LET
MLuttgens
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
: 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
>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
: 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
> 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
>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
<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
>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
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
:: 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
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
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
: 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
: 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
>: 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
>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
: 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
>
> 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
>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
: 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
: 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
: 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
: 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
>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
<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
:
>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
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
: 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
>
> 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
: It is interesting to note that the consequence of [Mluttgens'] demand that
: 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
>
> : "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
>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
>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
<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
<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
:: 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
>
> 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
>
> In article <38258056...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >
> >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.
> >
> >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.
> >
> >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
: 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
::: 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
: 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
>
> : "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
: "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
>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
>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
<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
>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
> >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
> >: 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 function, quadratic surface, quadrilateral... nope.
> > I don't recall the term. I'm going to presume it means "square". )
> >
> >: (It could be a cube, but I am satisfied with a 2d analysis.)
> >:
> >: L
> >: A----------B
> >: | |
> >: L| |
> >: | | -> v
> >: | |
> >: D----------C
> >
> >
> >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
: 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
: 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
>
> 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?
??
From where did I get that funny idea? :-)
a' = arcsin(L1'/L3') = arcsin(0.8575) = 121°
Paul
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Wed, 10 Nov 1999 23:39:18 +0100
>
>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.
>
Which Lorentz transform?
I used my transform, i.e. L'=L*sqrt(1-(v*cos(a))^2).
Before continuing this interesting discussion, I want to be sure
that you agree that an arm of length L making an angle a with
the velocity vector v is contracted according to my 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?
>
>From where did I get that funny idea? :-)
>
>a' = arcsin(L1'/L3') = arcsin(0.8575) = 121°
Yes, and according to the SR/LET angular transform
cos(a') = cos(a) * sqrt(1-v^2) / sqrt(1-(v*cos(a)^2)) = - 0.5145,
hence a' = 121°
Marcel Luttgens
MLuttgens
>Date : Thu, 11 Nov 1999 05:16:00 GMT
>
>: 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.
Do you disagree that the x,x'-projections must be identical
to the y or y'-projections in this special case of an angle a of 45° ?
Btw, do you agree now that an arm of length L, that makes an
angle a with the velocity vector v, is contracted according to
my transform L'=L*sqrt(1-(v*cos(a))^2) ?
[snip]
Marcel Luttgens
Paul B. Andersen
The transform which say that any distance vector between two
points on a rigid body (like an arm or a square or a circle),
with angle to the velocity a and length L, will transform thus:
Lx' = L*cos(a)*sqrt(1 - v^2)
Ly' = L*sin(a)
L' = L*sqrt(1-(v*cos(a))^2)
This is the transform you used above.
> I used my transform, i.e. L'=L*sqrt(1-(v*cos(a))^2).
No, you did not.
Marcel Luttgens 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).
Thus your transform is:
Lx' = L*sqrt(1-(v*cos(a))^2)*cos(a)
Ly' = L*sqrt(1-(v*cos(a))^2)*sin(a)
L' = L*sqrt(1-(v*cos(a))^2)
_This_ transform yields the rather interesting result:
L1' = 1, L1x' = 0, L1y' = 1, a' = a = 90°
L2' = 0.6, L2x' = 0.6, L2y' = 0, a' = a = 0°
L3' = 1.1662, L3x' = 0.82,L3y' = 0.82, a' = a = 135°
A-----B
A | |
* L3' |
L1'| * |
| * |
D-----C
L2'
Hard to figure out what kind of object this is, isn't it? :-)
The reason why you did not use this transformation is that
it is a huge self contradiction, which is impossible to use
for anything but one dimensional objects.
> Before continuing this interesting discussion, I want to be sure
> that you agree that an arm of length L making an angle a with
> the velocity vector v is contracted according to my transform.
I definitely do not agree to that.
It contracts according to the Lorentz transform as shown by you.
But isn't it settled now?
Your transformation is impossible.
So you ended up using the Lorentz contraction the correct way.
The way you called nosensical.
Paul
Wayne Throop
: Btw, do you agree now that an arm of length L, that makes an angle a
: with the velocity vector v, is contracted according to my transform
: L'=L*sqrt(1-(v*cos(a))^2) ?
I'm sure everybody can see why it is misleading to call this
a "transform" in this context. I'm sure MLuttgens can see it, too,
but he's TRYING to be misleading; that's the whole point.
The above applies to a measured angle a. It is incorrect for the coordinate angle a'. As I have said literally dozens of times now.
We can be absolutely sure MLuttgens already knew that, since he has
reponsed to many of the postings where I pointed it out before.
So why pretend it would be a ch ange for me to agree to
L'=L*sqrt(1-(v*cos(a))^2) ? Only to try to fool some readers.
I conclude he gets his jollies that way; lying and misleading folks.
: Do you disagree that the x,x'-projections must be identical to the y
: or y'-projections in this special case of an angle a of 45 ?
No. I just showed you that YOU don't think so, either.
When you actually worked a problem, you didn't set L1'=L2',
and L1' and L2' were x' and y' of the arm at measured 45 degrees.
And this is so bloody obvious, that there's no alternative to the
conclusion that MLuttgens is simply blustering to try to cover up the
fact that he knows very well that the angle is NOT invariant, and
in fact cannot consistently be invariant for any extended objects.
He knows very well that he's wrong,
but he's trying to see how many people he can fool.
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Fri, 12 Nov 1999 22:25:15 +0100
I proposed to use L' = L*sqrt(1-(v*cos(a))^2) on Jun 11,1999,
in "The MMX, a general analysis", where I showed that such
contraction explains the negative result of the MMX.
On Jul 11, 1999, in my article
<19990711043111...@ngol06.aol.com>,
"Re: A question to mathematicians and physicists", I wrote:
<<Without length contraction, the sides of triangle A are L*cos(a),
L*sin(a) and the hypotenuse L, where L is the length of the arm,
and a the angle between the arm and the velocity vector v
along the x-axis.
With length contraction, the hypotenuse becomes of course L',
and we get
L'^2 = (L*cos(a)*sqrt(1-v^2))^2 + (L*sin(a))^2), hence
L' = L * sqrt (1-(v*cos(a))^2) >>
So, I am well aware that L' can be derived from the contracted
x,x'-projection of the arm.
But a contracted projection is meaningless if the arm itself is not
contracted. And I claim that it is contracted according to
L' = L * sqrt (1-(v*cos(a))^2).
This doesn't contradict your above "Lorentz transform", if one
uses an angle a' instead of a to calculate the contracted arm
projections.
Indeed, L * sqrt (1-(v*cos(a))^2) * cos(a') = L * sqrt(1-v^2) * cos(a),
and L * sqrt (1-(v*cos(a))^2) * sin(a') = L * sin(a), if
cos(a') = sqrt(1-v^2) * cos(a) / sqrt (1-(v*cos(a))^2), and
sin(a') = sin(a) / sqrt (1-(v*cos(a))^2).
Note that I have used L' = L * sqrt (1-(v*cos(a))^2),
not L' = L * sqrt (1-(v*cos(a'))^2), to obtain the length of the
contracted arm.
Note also that using a and a' in the same formula is inconsistent.
When using cos(a) and sin(a) to calculate the projections of
the contracted arm, one gets indeed the results that you are
presenting above, but the "contradiction" is not "huger" than those
obtained by SR in the case of a rotating disk.
>> Before continuing this interesting discussion, I want to be sure
>> that you agree that an arm of length L making an angle a with
>> the velocity vector v is contracted according to my transform.
>
>I definitely do not agree to that.
>It contracts according to the Lorentz transform as shown by you.
>
Do you claim that L' is different from L*sqrt(1-(v*cos(a))^2) ?
>But isn't it settled now?
>Your transformation is impossible.
>So you ended up using the Lorentz contraction the correct way.
>The way you called nosensical.
>
>Paul
Marcel Luttgens
MLuttgens
>Date : Fri, 12 Nov 1999 21:47:49 GMT
>
>: mlut...@aol.com (MLuttgens)
>: Btw, do you agree now that an arm of length L, that makes an angle a
>: with the velocity vector v, is contracted according to my transform
>: L'=L*sqrt(1-(v*cos(a))^2) ?
>
>I'm sure everybody can see why it is misleading to call this
>a "transform" in this context. I'm sure MLuttgens can see it, too,
>but he's TRYING to be misleading; that's the whole point.
>
>The above applies to a measured angle a. It is incorrect for the coordinate
>angle a'. As I have said literally dozens of times now.
>
The arm is contracted by sqrt(1-(v*cos(a))^2), not by
sqrt(1-(v*cos(a'))^2).
>We can be absolutely sure MLuttgens already knew that, since he has
>reponsed to many of the postings where I pointed it out before.
>
>So why pretend it would be a change for me to agree to
>L'=L*sqrt(1-(v*cos(a))^2) ? Only to try to fool some readers.
>I conclude he gets his jollies that way; lying and misleading folks.
>
What is the correct formula giving the arm's contraction,
according to you?
Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
>: Do you disagree that the x,x'-projections must be identical to the y
>: or y'-projections in this special case of an angle a of 45 ?
>
>No.
Does your "No" mean that you agree that the projections are
identical?
> I just showed you that YOU don't think so, either.
>When you actually worked a problem, you didn't set L1'=L2',
>and L1' and L2' were x' and y' of the arm at measured 45 degrees.
>
>And this is so bloody obvious, that there's no alternative to the
>conclusion that MLuttgens is simply blustering to try to cover up the
>fact that he knows very well that the angle is NOT invariant, and
>in fact cannot consistently be invariant for any extended objects.
>
The invariance of the angle is comptatible with the negative result
of the MMX.
(see The MMX, a general analysis (Ver. 5)).
>He knows very well that he's wrong,
>but he's trying to see how many people he can fool.
>
You have funny ideas.
In fact, you don't even grasp the implications of your own derivations.
For instance, if the arm makes an angle a with the velocity vector v,
the "maxwellian" ether times are given by
T(a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2).
Two different contractions factors can then be applied to the arm:
1) sqrt(1-(v*cos(a))^2)
This is my factor, and also yours (and that of Paul B. Andersen).
If one considers that the coordinate angle a' is identical to the
measured angle a, the "relativistic" ether times is given by
T(a)' = (2*L*sqrt(1-(v*cos(a))^2)/(1-v^2)) * sqrt(1-(v*sin(a))^2),
i.e. my formula, that correctly explains the null MMX result.
But if the angle becomes a' in sqrt(1-(v*sin(a))^2), the resulting
relativistic formula
T(a)' = (2*L*sqrt(1-(v*cos(a))^2)/(1-v^2)) * sqrt(1-(v*sin(a'))^2)
becomes
T(a)' = 2L / sqrt(1-v^2), which is independent of any angle
(your formula, which is also compatible with the MMX).
This is a direct consequence of your angular transform
sin (a') = sin(a) / sqrt(1-(v*cos(a)^2).
2) sqrt(1-v^2)
This factor is used in the SR formula giving the effect of uniform
motion upon the field of a point electric charge.
Such formula contains the expression
L' = L * sqrt(1 - (v*sin(a))^2)) / sqrt(1-v^2), where L is the distance
of a point P of the field to a point charge q (the charge is moving
at v relative to the frame S, and stationary relative to S').
Indeed, T(a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2) then gives
T(a)' = (2*L*sqrt(1-v^2)/(1-v^2)) * sqrt(1-(v*sin(a))^2)
= 2*L*sqrt(1-(v*sin(a))^2) / sqrt(1-v^2)
In the SR derivation of this formula, the angle a has been
considered as invariant, in contradiction with your claim that
the angle "cannot consistently be invariant for any extended
objects".
>
>Wayne Throop
Marcel Luttgens
Wayne Throop
: But a contracted projection is meaningless if the arm itself is not
: contracted. And I claim that it is contracted according to L' = L *
: sqrt (1-(v*cos(a))^2).
Of course. But that's not controversial. What's controversial
is MLuttgens' bizarre insistance that the angle is invariant.
So it's pretty dishonest of him to keep going over and over
things long agreed upon, while pointedly ignoring the fact
that the angle is NOT invariant.
Especially dishonest given the clear and simple demonstration
that 1. the method he claims to use, he does not use, and 2. the
method he claims to use is, in fact, self-contradictory.
: Note also that using a and a' in the same formula is inconsistent.
Why? For example, what's wrong with L*sqrt(1-(v*cos(a))^2)*cos(a') ?
After all, L'=L*sqrt(1-(v*cos(a))^2), so that's the same as L'*cos(a').
What's wrong with that?
I expect MLuttgens, being a troll, won't be honest here,
so a small hint: nothing's wrong with it.
: Do you claim that L' is different from L*sqrt(1-(v*cos(a))^2) ?
No. Do you still claim that a=a'?
Wayne Throop
:: The above applies to a measured angle a. It is incorrect for the
:: coordinate angle a'. As I have said literally dozens of times now.
: mlut...@aol.com (MLuttgens)
: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
: sqrt(1-(v*cos(a'))^2).
That IS what I just said. For about the 37th time. Twit.
: What is the correct formula giving the arm's contraction, according to
: you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
That IS what I just said. For about the 38th time. Twit.
: You have funny ideas.
You have self-contradictory ideas. Or rather,
you dishonestly fake having self-contradictory ideas,
knowing very well that you are spouting falsehoods.
: This is a direct consequence of your angular transform sin (a') =
: sin(a) / sqrt(1-(v*cos(a)^2).
Which, in turn, is a direct consequence of LET.
As you know very well. Dishonest twit.
: In the SR derivation of this formula, the angle a has been considered
: as invariant, in contradiction with your claim that the angle "cannot
: consistently be invariant for any extended objects".
Exactly. That's why your alleged "SR derivation" is neither SR,
nor a valid derivation at all. As you know very well. Twit.
Paul B. Andersen
Which has not been disputed.
That does follow from the Lorentz transform.
> This doesn't contradict your above "Lorentz transform", if one
> uses an angle a' instead of a to calculate the contracted arm
> projections.
>
> Indeed, L * sqrt (1-(v*cos(a))^2) * cos(a') = L * sqrt(1-v^2) * cos(a),
> and L * sqrt (1-(v*cos(a))^2) * sin(a') = L * sin(a), if
> cos(a') = sqrt(1-v^2) * cos(a) / sqrt (1-(v*cos(a))^2), and
> sin(a') = sin(a) / sqrt (1-(v*cos(a))^2).
> Note that I have used L' = L * sqrt (1-(v*cos(a))^2),
> not L' = L * sqrt (1-(v*cos(a'))^2), to obtain the length of the
> contracted arm.
> Note also that using a and a' in the same formula is inconsistent.
And you want us _not_ to note that you insisted:
"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)."
The implication of which is that a' = a ?
So you finally admit that your "transform" is self contradictory.
(But not "huge" of course. :-) )
> >> Before continuing this interesting discussion, I want to be sure
> >> that you agree that an arm of length L making an angle a with
> >> the velocity vector v is contracted according to my transform.
> >
> >I definitely do not agree to that.
> >It contracts according to the Lorentz transform as shown by you.
> >
>
> Do you claim that L' is different from L*sqrt(1-(v*cos(a))^2) ?
What is the point with this, Marcel?
Do you wish to make the impression that we ever claimed that,
in order to make it appear that _you_ were right? :-)
Look up Wayne's or mine derivation of the length, and see
what you find there.
You know what the dispute is about, Marcel.
It is the following statement of yours:
| 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).
And you even wrote:
| A new SR/GR should be based on the contraction factor
| sqrt(1-(v*cos(a))^2), and the constancy of the angle a.
And now you seem to wish to make it appear that you all the time
have said that the rod contracts in the way you above call
"silly idea" and "nonsensical"? :-)
> >But isn't it settled now?
> >Your transformation is impossible.
> >So you ended up using the Lorentz contraction the correct way.
> >The way you called nosensical.
Isn't it?
And the statement in the subject field is false, isn't it?
Paul
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Mon, 15 Nov 1999 23:59:52 +0100
>
>MLuttgens wrote:
>>
[snip]
>> But a contracted projection is meaningless if the arm itself is not
>> contracted. And I claim that it is contracted according to
>> L' = L * sqrt (1-(v*cos(a))^2).
>
>Which has not been disputed.
>That does follow from the Lorentz transform.
>
And also from my general analysis of the MMX.
>> This doesn't contradict your above "Lorentz transform", if one
>> uses an angle a' instead of a to calculate the contracted arm
>> projections.
>>
>> Indeed, L * sqrt (1-(v*cos(a))^2) * cos(a') = L * sqrt(1-v^2) * cos(a),
>> and L * sqrt (1-(v*cos(a))^2) * sin(a') = L * sin(a), if
>> cos(a') = sqrt(1-v^2) * cos(a) / sqrt (1-(v*cos(a))^2), and
>> sin(a') = sin(a) / sqrt (1-(v*cos(a))^2).
>> Note that I have used L' = L * sqrt (1-(v*cos(a))^2),
>> not L' = L * sqrt (1-(v*cos(a'))^2), to obtain the length of the
>> contracted arm.
>> Note also that using a and a' in the same formula is inconsistent.
>
>And you want us _not_ to note that you insisted:
> "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)."
>The implication of which is that a' = a ?
>
Yes.
No, I wanted to stress the huge SR contradictions. :-)
Nobody knows how L could physically contracts to L'. There is
no known reason why a physical object, e.g. a triangle, should
retain its form after such contraction.
But, in fact, the contraction is probably not physical at all;
I often said, it is a mere mathematical artefact. So, why should
I interpret the above calculated projections as contradictory?
I maintain that position.
>And now you seem to wish to make it appear that you all the time
>have said that the rod contracts in the way you above call
>"silly idea" and "nonsensical"? :-)
>
????
>> >But isn't it settled now?
>> >Your transformation is impossible.
What about the SR contractions of the rotating disk?
Are they possible?
>> >So you ended up using the Lorentz contraction the correct way.
>> >The way you called nosensical.
>
>Isn't it?
>
No.
>And the statement in the subject field is false, isn't it?
>
>Paul
Marcel Luttgens
MLuttgens
>Date : Mon, 15 Nov 1999 21:48:05 GMT
>
>::: L'=L*sqrt(1-(v*cos(a))^2) ?
>
>:: The above applies to a measured angle a. It is incorrect for the
>:: coordinate angle a'. As I have said literally dozens of times now.
>
>: mlut...@aol.com (MLuttgens)
>: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
>: sqrt(1-(v*cos(a'))^2).
>
>That IS what I just said. For about the 37th time. Twit.
>
>: What is the correct formula giving the arm's contraction, according to
>: you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
>
>That IS what I just said. For about the 38th time. Twit.
>
No, it can only be L'=L*sqrt(1-(v*cos(a))^2).
Using a' gives wrong results for the length L' of the contracted arm.
I wrote:
If the arm makes an angle a with the velocity vector v,
the "maxwellian" ether times are given by
T(a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2).
Two different contractions factors can then be applied to the arm:
1) sqrt(1-(v*cos(a))^2)
This is my factor, and also yours (and that of Paul B. Andersen).
If one considers that the coordinate angle a' is identical to the
measured angle a, the "relativistic" ether times is given by
T(a)' = (2*L*sqrt(1-(v*cos(a))^2)/(1-v^2)) * sqrt(1-(v*sin(a))^2),
i.e. my formula, that correctly explains the null MMX result.
But if the angle becomes a' in sqrt(1-(v*sin(a))^2), the resulting
relativistic formula
T(a)' = (2*L*sqrt(1-(v*cos(a))^2)/(1-v^2)) * sqrt(1-(v*sin(a'))^2)
becomes
T(a)' = 2L / sqrt(1-v^2), which is independent of any angle
(your formula, which is also compatible with the MMX).
This is a direct consequence of your angular transform
sin (a') = sin(a) / sqrt(1-(v*cos(a)^2).
Your comment:
Which, in turn, is a direct consequence of LET.
My reply:
How could LET be generally applicable?
True, if a=0°, sin(a)=0 and cos(a)=1, thus sin(a')=0, a'=a, hence
LET gives the same result T(a)' = 2L / sqrt(1-v^2) as my formula
T(a)' = (2*L*sqrt(1-(v*cos(a))^2)/(1-v^2)) * sqrt(1-(v*sin(a'))^2)
If a=90°, sin(a)=1 and cos(a)=0, thus sin (a') = sin(a) /
sqrt(1-(v*cos(a)^2) = 1, a'=a. Again, LET and my formula give
identical results, i.e. T(a)' = 2L / sqrt(1-v^2).
It is only when a<>0 or a<>90° that the results are slightly different,
but LET/SR have limited themselves to the special cases 0° and
90°, so they cannot make any sensible prediction for the general
case.To realize this, just consider the Lorentz transformation
x'=gamma(x-vt), y'=y.
2) sqrt(1-v^2)
This factor is used in the SR formula giving the effect of uniform
motion upon the field of a point electric charge.
Such formula contains the expression
L' = L * sqrt(1 - (v*sin(a))^2)) / sqrt(1-v^2), where L is the distance
of a point P of the field to a point charge q (the charge is moving
at v relative to the frame S, and stationary relative to S').
Indeed, T(a) = (2*L/(1-v^2)) * sqrt(1-(v*sin(a))^2) then gives
T(a)' = (2*L*sqrt(1-v^2)/(1-v^2)) * sqrt(1-(v*sin(a))^2)
= 2*L*sqrt(1-(v*sin(a))^2) / sqrt(1-v^2)
In the SR derivation of this formula, the angle a has been
considered as invariant, in contradiction with your claim that
the angle "cannot consistently be invariant for any extended
objects".
Your comment:
Exactly. That's why your alleged "SR derivation" is neither SR,
nor a valid derivation at all.
My reply:
It is not my SR derivation. You can find that derivation in any book
giving the SR formula for the effect of uniform motion upon the
field of a point electric charge.
>Wayne Throop
Marcel Luttgens
Paul B. Andersen
|MLuttgens wrote:
|: But a contracted projection is meaningless if the arm itself is not
|: contracted. And I claim that it is contracted according to L' = L *
|: sqrt (1-(v*cos(a))^2).
|
|| Of course. But that's not controversial. What's controversial
|| is MLuttgens' bizarre insistance that the angle is invariant.
|| So it's pretty dishonest of him to keep going over and over
|| things long agreed upon, while pointedly ignoring the fact
|| that the angle is NOT invariant.
So Wayne Throop say : L' = L*sqrt(1-(v*cos(a))^2)
(And he never said otherwise.)
|MLuttgens repeats:
|: Btw, do you agree now that an arm of length L, that makes an
|: angle a with the velocity vector v, is contracted according to
|: my transform L'=L*sqrt(1-(v*cos(a))^2) ?
|
|Wayne Throop replies:
|| The above applies to a measured angle a. It is incorrect for the
|| coordinate angle a'. As I have said literally dozens of times now.
|| We can be absolutely sure MLuttgens already knew that, since he has
|| reponsed to many of the postings where I pointed it out before.
|| So why pretend it would be a change for me to agree to
|| L'=L*sqrt(1-(v*cos(a))^2) ? Only to try to fool some readers.
|| I conclude he gets his jollies that way; lying and misleading folks.
So Wayne Throop again say: L' = L*sqrt (1-(v*cos(a))^2)
| MLuttgens keep repeating:
|: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
|: sqrt(1-(v*cos(a'))^2).
|
|Wayne Throop replies:
|| That IS what I just said. For about the 37th time. Twit.
So Wayne Throop again say: L' = L*sqrt (1-(v*cos(a))^2)
| MLuttgens gets a bright idea, and writes:
|: What is the correct formula giving the arm's contraction, according to
|: you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
|
| Wayne Throop does not note that MLuttgens has sneaked in an ' in
| the question, and replies:
|| That IS what I just said. For about the 38th time. Twit.
And then MLuttgens ignore that the question has been answered
over and over and over again, and pretend to believe that this
last obvious accident is Wayne's only answer.
And then he goes off.
Marcel, please stop this nonsense!
You know very well that the dispute is not and never was
about the equation: L' = L*sqrt (1-(v*cos(a))^2),
so why do you pretend that it is?
The dispute is about whether the angle of the arm is invariant or not.
You insisted that it is.
Wayne and I insisted it is not.
You have now realized that you were wrong, and that your claim
is self contradictory.
So why don't you admit that instead of this disgraceful behaviour?
Isn't this below your dignity?
Paul
Wayne Throop
::::: L'=L*sqrt(1-(v*cos(a))^2) ?
:::: The above applies to a measured angle a. It is incorrect for the
:::: coordinate angle a'. As I have said literally dozens of times now.
::: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
::: sqrt(1-(v*cos(a'))^2).
::: What is the correct formula giving the arm's contraction, according
::: to you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
Can MLuttgens read plain english? Aparently not.
:: That IS what I just said. For about the 37th time. Twit.
::: What is the correct formula giving the arm's contraction, according
::: to you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
:: That IS what I just said. For about the 38th time. Twit.
( It would have been more apt to say "What did I just say",
given MLuttgens' pretended misunderstanding below...
still makes him a twit, though. )
: Using a' gives wrong results for the length L' of the contracted arm.
The length of the arm in a' is sqrt((1-v^2)/(1-(v*sin(a'))^2))
The length of the arm in a is sqrt(1-(v*cos(a))^2)
I first pointed this out in <9341...@sheol.org>, in July of 1999.
Since then, I've stated many times what the length is.
I even posted the derivation in a several times, since MLuttgens
kept asking about it. Eg, in the above post.
I keep pointing out that I AGREE L'=sqrt(1-(v*cos(a))^2),
but DISagree that a=a'. Indeed, quoting an earlier article,
MLuttgens acknowledges
: 1) sqrt(1-(v*cos(a))^2)
: This is my factor, and also yours (and that of Paul B. Andersen).
So why is he asking what I think the length is? I conclude it's because
he's trying to distract attention from the fact that the length has
never been an issue. It is only the invariance of the angle that has
been at issue.
But even the invariance of the angle has not HONESTLY been at issue.
MLuttgens knows very well that a does not equal a'.
MLuttgens himself worked the problem that SHOWS a is not equal to a'.
In article <19991110082306...@ngol06.aol.com> .
He shows clearly that for a diagonal arm (a>0 degrees and a<90 degrees),
a does not equal a'. So he knows very well that his claim about a=a'
is false. But he continues to blow smoke, trying to pretend there's
some disagreement about L', and trying to pretend there's even a
self-consistent possibility, that a=a'.
See, he keeps pretending:
: If one considers that the coordinate angle a' is identical to the
: measured angle a, [...]
all the while knowing very well that a does NOT equal a'.
To continue to pretend it does, or even MIGHT do so, is dishonest.
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Wed, 17 Nov 1999 16:24:59 +0100
>
>
>|MLuttgens wrote:
>|: But a contracted projection is meaningless if the arm itself is not
>|: contracted. And I claim that it is contracted according to L' = L *
>|: sqrt (1-(v*cos(a))^2).
>|
>| Wayne Throop replies:
>|| Of course. But that's not controversial. What's controversial
>|| is MLuttgens' bizarre insistance that the angle is invariant.
>|| So it's pretty dishonest of him to keep going over and over
>|| things long agreed upon, while pointedly ignoring the fact
>|| that the angle is NOT invariant.
>
>So Wayne Throop say : L' = L*sqrt(1-(v*cos(a))^2)
>(And he never said otherwise.)
>
>|MLuttgens repeats:
>|: Btw, do you agree now that an arm of length L, that makes an
>|: angle a with the velocity vector v, is contracted according to
>|: my transform L'=L*sqrt(1-(v*cos(a))^2) ?
>|
>|Wayne Throop replies:
>|| The above applies to a measured angle a. It is incorrect for the
>|| coordinate angle a'. As I have said literally dozens of times now.
>|| We can be absolutely sure MLuttgens already knew that, since he has
>|| reponsed to many of the postings where I pointed it out before.
>|| So why pretend it would be a change for me to agree to
>|| L'=L*sqrt(1-(v*cos(a))^2) ? Only to try to fool some readers.
>|| I conclude he gets his jollies that way; lying and misleading folks.
>
>So Wayne Throop again say: L' = L*sqrt (1-(v*cos(a))^2)
>
>| MLuttgens keep repeating:
>|: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
>|: sqrt(1-(v*cos(a'))^2).
>|
>|Wayne Throop replies:
>|| That IS what I just said. For about the 37th time. Twit.
>
>So Wayne Throop again say: L' = L*sqrt (1-(v*cos(a))^2)
>
>| MLuttgens gets a bright idea, and writes:
>|: What is the correct formula giving the arm's contraction, according to
>|: you? Is it L'=L*sqrt(1-(v*cos(a'))^2) ?
>|
>| Wayne Throop does not note that MLuttgens has sneaked in an ' in
>| the question, and replies:
>|| That IS what I just said. For about the 38th time. Twit.
>
>And then MLuttgens ignore that the question has been answered
>over and over and over again, and pretend to believe that this
>last obvious accident is Wayne's only answer.
>
Are you so sure that Wayne Throop missed the ' in the formula?
For instance, on June 25, 1999, he wrote in "Re: An insoluble
problem for Wayne Throop (and all SRians)":
< [...] there is no such thing as length contraction or time dilation by
gamma(v'), since nothing at all is moving at v'. >
Note that v' is the component of the velocity vector along the arm,
and its value is given by v' = v*cos(a).
>And then he goes off.
>
> Marcel, please stop this nonsense!
>
> You know very well that the dispute is not and never was
> about the equation: L' = L*sqrt (1-(v*cos(a))^2),
> so why do you pretend that it is?
>
I don't know "very well", because Wayne Throop is accustomed
to cleverly mix white and black in the same sentence, so I am
not always sure of what he really means. Ambiguity is one of his
favorite debating tactics.
> The dispute is about whether the angle of the arm is invariant or not.
> You insisted that it is.
> Wayne and I insisted it is not.
> You have now realized that you were wrong, and that your claim
> is self contradictory.
>
How can you seriously pretend that I have now realized being
wrong? I would like to say that the use of such gratuitous assertions
belongs to *your* favorite debating tactic.
> So why don't you admit that instead of this disgraceful behaviour?
> Isn't this below your dignity?
>
Why don't you admit that you have no serious clue that the angle
is not invariant. You just rely on the Lorentz/SR transform, that is only
relevant for the degenerate cases 0° and 90°.
And don't forget that my formula gives exactly the same results
as Lorentz/SR ones in those cases.
>Paul
Marcel Luttgens
Paul B. Andersen
I figured as much. We _did_ notice, however.
Sure I understand that you wanted to divert the attention by
referring to an off topic issue where _you_ think SR is
self contradictory.
But that does change the fact that your statement implies
an admission of that your "transformation" is self contradictory.
You are effectually saying:
"Yes, I admit that my "transformation" yields self contradictory
results and thus must be self contradictory, but I think SR give
a self contradictory prediction in another case."
Even if the latter had been true, your transformation would
still be self contradictory, no?
> Nobody knows how L could physically contracts to L'.
Well, the etherists would't agree. :-)
But I simply say: the rod does not "physically contract".
> There is
> no known reason why a physical object, e.g. a triangle, should
> retain its form after such contraction.
It doesn't, according to LET.
> But, in fact, the contraction is probably not physical at all;
> I often said, it is a mere mathematical artefact.
According to SR, the contraction is a "mathematical artefact"
in exactly the same way as speed is a "mathematical artefact".
"Speed" is no intrinsic property of the object, it is a relationship
between the frame of reference and the object. The fact that the
speed is different in different frames does not mean that "something
is happening to the object". Same thing with the length of the rod.
The length of the rod is different in different frames of reference,
but that does not mean that anything is happening to the rod.
Rather obvious, really. The length and speed of the same single rod
can at the same time be measured in a number of different frames
of reference, having different length and speed in each frame.
Being observed in different frames of reference have no physical
consequence for the rod, but the different states of motion of
the frames of reference do indeed have a physical consequence
for the measured length and speed.
The length and the speed of the object can be measured with real,
physical instruments. Personally, I tend to call measurable
entities for "real".
But if you insist on calling them "mathematical artifacts"
because the measured values depend on the state of motion
of the frame of reference, by all means - do so.
And the next time you are caught speeding, you can try to
tell the policeman that the speed he has measured is a mathematical
artefact of his chosen frame of reference. Tell him that
the speed zero is exactly equally real as the speed he measured.
You may even succeed to convince him, in which case you will
be fined for your "mathematically artificial speed in the ground
frame".
> So, why should
> I interpret the above calculated projections as contradictory?
Come on, is nothing obvious to you?
Your "transform" is mathematically inconsistent, and thus
it cannot even be a "mathematical artifact".
Your "transform" implies that each point on the body
at the same time is at an infinite number of different
positions!
In the example above (repeated below), I demonstrated that for
the point A.
|_This_ transform yields the rather interesting result:
| L1' = 1, L1x' = 0, L1y' = 1, a' = a = 90°
| L2' = 0.6, L2x' = 0.6, L2y' = 0, a' = a = 0°
| L3' = 1.1662, L3x' = 0.82,L3y' = 0.82, a' = a = 135°
|
| A-----B
| A | |
| * L3' |
| L1'| * |
| | * |
| D-----C
| L2'
|
In this case, A has to be at two different positions at
the same time.
But you can calculate as many different positions you want
by going around different triangles.
And you can do the same for any other point on the body.
It is a _huge_ self contradiction.
And your admission of that my calculations with self contradictory
result is correct, is of course an admission of that the rules
by which the calculations were done must be self contradictory.
And you admit it is, and still you "maintain your position"?
I find that unbelievable.
So what would we call a person who claim to believe what he
admits is impossible?
> >And now you seem to wish to make it appear that you all the time
> >have said that the rod contracts in the way you above call
> >"silly idea" and "nonsensical"? :-)
> >
>
> ????
>
> >> >But isn't it settled now?
> >> >Your transformation is impossible.
>
> What about the SR contractions of the rotating disk?
> Are they possible?
Yes.
You may find them weird, but they are mathematically
consistent.
> >> >So you ended up using the Lorentz contraction the correct way.
> >> >The way you called nosensical.
> >
> >Isn't it?
> >
>
> No.
Well. If you keep insisting that the "transformation" you
have admitted yields self contradictory results, is right,
then I suppose any further arguments will be futile.
"Is too!" arguments cannot be refuted by logic.
Paul
Wayne Throop
: Are you so sure that Wayne Throop missed the ' in the formula?
: For instance, on June 25, 1999, he wrote in "Re: An insoluble problem
: for Wayne Throop (and all SRians)": < [...] there is no such thing as
: length contraction or time dilation by gamma(v'), since nothing at all
: is moving at v'. > Note that v' is the component of the velocity
: vector along the arm, and its value is given by v' = v*cos(a).
And sure enough, there isn't. The contraction is by the factor
sqrt(1-v^2), not sqrt(1-v'^2) (where v' is not velocity in the primed
frame, but a quantity described by MLuttgens as cos(a)*v).
Because, of course, technically speaking, the rod's "length" is not
what is contracted. Contraction in LET and SR is always by a factor
of sqrt(1-v^2), and never any other. But it is only x-extent that
is co ntracted.
Now, MLuttggens knows all this perfectly well.
So he remains a major hypocrite to pretend my position is at all
inconsistent or unclear. The length of the rod in coordinates in which
it moves at v, compared to the length of the same rod in coordinates in
which it does not move.
MLuttgens knows perfectly well that that's what I've *ALWAYS*
said, and so his trying to kick up dust and obfuscation can only
mean he's trying to confuse the issue. Intentionally.
: I don't know "very well", because Wayne Throop is accustomed to
: cleverly mix white and black in the same sentence,
Where? In the above, I stated qu ite clearly what I meant,
and added lots of words to make clear what I meant by "length contraction",
and why the ratio of rod lengths isn't a length contraction.
MLuttgens k nows all this, but pretends I was unclear.
: Why don't you admit that you have no serious clue that the angle is
: not invariant.
I showed you that it is not. You yourself have shown that it is not.
This pretense is very thin, MLuttgens; you will have to do better
if you hope to fool more of the people more of the time.
Paul B. Andersen
>
> In article <3832C8CB...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >
> >
> >|MLuttgens wrote:
> >|: contracted. And I claim that it is contracted according to L' = L *
> >|: sqrt (1-(v*cos(a))^2).
> >|
> >|| Of course. But that's not controversial. What's controversial
> >|| is MLuttgens' bizarre insistance that the angle is invariant.
> >|| So it's pretty dishonest of him to keep going over and over
> >|| things long agreed upon, while pointedly ignoring the fact
> >|| that the angle is NOT invariant.
> >
> >So Wayne Throop say : L' = L*sqrt(1-(v*cos(a))^2)
> >(And he never said otherwise.)
> >
> >|MLuttgens repeats:
> >|: angle a with the velocity vector v, is contracted according to
>
> For instance, on June 25, 1999, he wrote in "Re: An insoluble
> problem for Wayne Throop (and all SRians)":
> < [...] there is no such thing as length contraction or time dilation by
> gamma(v'), since nothing at all is moving at v'. >
> Note that v' is the component of the velocity vector along the arm,
> and its value is given by v' = v*cos(a).
And? Isn't what Wayne writes rather obvious?
> >And then he goes off.
> >
> > Marcel, please stop this nonsense!
> >
> > You know very well that the dispute is not and never was
> > about the equation: L' = L*sqrt (1-(v*cos(a))^2),
> > so why do you pretend that it is?
> >
>
> I don't know "very well", because Wayne Throop is accustomed
> to cleverly mix white and black in the same sentence, so I am
> not always sure of what he really means. Ambiguity is one of his
> favorite debating tactics.
Of course you know very well indeed.
And we know you know.
You are not fooling anyone.
> > The dispute is about whether the angle of the arm is invariant or not.
> > You insisted that it is.
> > Wayne and I insisted it is not.
> > You have now realized that you were wrong, and that your claim
> > is self contradictory.
> >
>
> How can you seriously pretend that I have now realized being
> wrong? I would like to say that the use of such gratuitous assertions
> belongs to *your* favorite debating tactic.
Because I assume that when someone proposes a "theory", which is proven
to be inconsistent, then that someone will realize that the proposed
"theory" is wrong.
I cannot bring myself to "seriously pretend" that you have not
realized that the "transform" which you admit produce self
contradictory results, is wrong.
> > So why don't you admit that instead of this disgraceful behaviour?
> > Isn't this below your dignity?
> Why don't you admit that you have no serious clue that the angle
> is not invariant. You just rely on the Lorentz/SR transform, that is only
> relevant for the degenerate cases 0° and 90°.
???
The angle is not invariant because the real world is Lorentz
invariant.
But you are right, I have indeed no serious clue as to why
Nature is as it is. But it is. Isn't it?
> And don't forget that my formula gives exactly the same results
> as Lorentz/SR ones in those cases.
Your "formula" is a huge self contradiction.
You have indeed realized that.
So why do you not admit it?
Paul
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Sat, 20 Nov 1999 11:45:21 +0100
>
>MLuttgens wrote:
[snip]
>> Are you so sure that Wayne Throop missed the ' in the formula?
>>
>> For instance, on June 25, 1999, he wrote in "Re: An insoluble
>> problem for Wayne Throop (and all SRians)":
>> < [...] there is no such thing as length contraction or time dilation by
>> gamma(v'), since nothing at all is moving at v'. >
>> Note that v' is the component of the velocity vector along the arm,
>> and its value is given by v' = v*cos(a).
>
>And? Isn't what Wayne writes rather obvious?
>
MLuttgens:
The arm is contracted by sqrt(1-(v*cos(a))^2), not by sqrt(1-(v*cos(a'))^2).
Wayne Throop replies:
That IS what I just said. For about the 37th time
But Wayne Throop said in another post:
... there is no such thing as length contraction by sqrt(1-(v*cos(a))^2).
I find this obviously contradictory, despite any slim, slick and
crooked counter argumentation.
[snip]
>> Why don't you admit that you have no serious clue that the angle
>> is not invariant. You just rely on the Lorentz/SR transform, that is only
>> relevant for the degenerate cases 0° and 90°.
>
>???
>The angle is not invariant because the real world is Lorentz
>invariant.
>
>But you are right, I have indeed no serious clue as to why
>Nature is as it is. But it is. Isn't it?
>
Nature is not illogical, unlike your demonstration of the non-
invariance of the angle a:
Paraphrase of Throop/Andersen's demonstration that a'<>a
BEGIN
Let's consider an interferometer with perpendicular arms of
length L.
If the interferometer is moving with speed v, the length of
the traverse y arm is not shortened, it is L. But since the
arm is moving, the length of the light path back and forth
the moving arm will be: L/sqrt(1 - v^2/c^2).
The length of the parallel x-arm is shortened to L*sqrt(1 - v^/c^2).
The length of the light path back and forth along this arm will be:
L*sqrt(1 - v^/c^2)*c/(c-v) + L*sqrt(1 - v^/c^2)*c/(c+v) =
L/sqrt(1 - v^/c^2).
Thus the light paths along the two arms are equal; the light
will be in phase - no fringe shift.
Now let's consider a third arm making an angle of a=45° with
the velocity vector v, and whose rest length L3 is L/cos(45).
The lengths of its x,x' and y'-projections are of course
L3*cos(45) and L3*sin(45) = L.
As the parallel arm of length L is shortened to L*sqrt(1 - v^/c^2),
the x,x'-projection of L3 becomes L3*cos(45)*sqrt(1 - v^/c^2),
but its y'-projection remains L3*sin(45).
So, the length L3' of the moving arm is given by
sqrt((L3*cos(45)*sqrt(1-v^2/c^2))^2 + (L3*sin(45))^2) =
L3*sqrt(1-(cos(45)*v/c)^2). Iow, the 45°-arm is apparently
shortened by sqrt(1-(cos(45)*v/c)^2), or, more generally,
by sqrt(1-(cos(a)*v/c)^2).
But the x,x'-projection of the moving arm, i.e.
L3' * cos(a) = L3*sqrt(1-(cos(45)*v/c)^2) * cos(45), is
different from L3*cos(45)*sqrt(1 - v^/c^2), and its y'-projection
L3' * sin(a) = L3*sqrt(1-(cos(45)*v/c)^2) * sin(45) is
different from L3*sin(45), hence the moving arm must make
with the velocity vector an angle a' which is different from
a=45°.
It is easy to show that the required angular transformations are
sin (a') = sin(a) / sqrt(1-(v*cos(a)^2)
cos(a') = cos(a) * sqrt(1-v^2) / sqrt(1-(v*cos(a)^2))
END
But it is also easy to see that this Throop/Andersen's demonstration
of " 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" is logically wrong.
>> And don't forget that my formula gives exactly the same results
>> as Lorentz/SR ones in those cases.
>
>Your "formula" is a huge self contradiction.
>You have indeed realized that.
>So why do you not admit it?
>
I don't have to admit your false claims. But why don't you admit
that your demonstration of the non-invariance of the angle a
is logically inconsistent?
>Paul
Marcel Luttgens
Wayne Throop
::: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
::: sqrt(1-(v*cos(a'))^2).
:: thr...@sheol.org (Wayne Throop)
:: That IS what I just said. For about the 37th time
: But Wayne Throop said in another post
:: there is no such thing as length contraction by sqrt(1-(v*cos(a))^2)
: mlut...@aol.com (MLuttgens)
: I find this obviously contradictory, despite any slim, slick and
: crooked counter argumentation.
They MLuttgens is being dishonest. He knows very well, because I have
explained in extreme detail in postings he has responded to, exactly
what I meant by those two statements that you have snipped from context.
For MLuttgens to pretend to find them contradictory at this point is
ridiculous, hence I conclude, dishonest.
The term "length contraction" refers to an anisotropic effect in LET.
It always occurs by a factor of sqrt(1-v^2).
In LET, there is NO "length contraction" which has BOTH a ratio of
sqrt(1-(v*cos(a))^2) AND preserves the angle a. (0 < a < 90)
All of which has been discussed to death for the last two months.
For MLuttgens to claim he didn't know what I meant at this point...
well, "dishonest" doesn't even half cover it.
: Nature is not illogical, unlike your demonstration of the non-
: invariance of the angle a:
What about YOUR OWN demonstration of the change in the angle between frames?
Is it illogical, too? Where do you claim you make your mistake?
: I don't have to admit your false claims.
You don't have to admit true ones, either, as you have demonstrated.
Paul B. Andersen
>
> In article <38367BC1...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Sat, 20 Nov 1999 11:45:21 +0100
> >
> >MLuttgens wrote:
>
> [snip]
>
> >> Are you so sure that Wayne Throop missed the ' in the formula?
> >>
> >> For instance, on June 25, 1999, he wrote in "Re: An insoluble
> >> problem for Wayne Throop (and all SRians)":
> >> < [...] there is no such thing as length contraction or time dilation by
> >> gamma(v'), since nothing at all is moving at v'. >
> >> Note that v' is the component of the velocity vector along the arm,
> >> and its value is given by v' = v*cos(a).
> >
> >And? Isn't what Wayne writes rather obvious?
I suppose you must agree, since you instead of commenting on
the quoted text above, select a quite different one:
> MLuttgens:
> The arm is contracted by sqrt(1-(v*cos(a))^2), not by sqrt(1-(v*cos(a'))^2).
> Wayne Throop replies:
> That IS what I just said. For about the 37th time.
> But Wayne Throop said in another post:
> ... there is no such thing as length contraction by sqrt(1-(v*cos(a))^2).
>
> I find this obviously contradictory, despite any slim, slick and
> crooked counter argumentation.
You are not fooling anyone by your out of context quotation.
You know perfectly well that the dispute is about the invariance
of the angle, not about the length of the moving arm.
> [snip]
>
> >> Why don't you admit that you have no serious clue that the angle
> >> is not invariant. You just rely on the Lorentz/SR transform, that is only
> >> relevant for the degenerate cases 0° and 90°.
> >
> >???
> >The angle is not invariant because the real world is Lorentz
> >invariant.
> >
> >But you are right, I have indeed no serious clue as to why
> >Nature is as it is. But it is. Isn't it?
> >
>
> Nature is not illogical, unlike your demonstration of the non-
> invariance of the angle a:
>
It is?
Why don't you show us the proof of that, then.
Why don't you show us how - say a square - transforms according
to the Lorentz transform, and show us that it is self contradictory?
> >> And don't forget that my formula gives exactly the same results
> >> as Lorentz/SR ones in those cases.
> >
> >Your "formula" is a huge self contradiction.
> >You have indeed realized that.
> >So why do you not admit it?
> >
>
> I don't have to admit your false claims.
Well, I showed you your "transform" applied on the rigid body:
|_This_ transform yields the rather interesting result:
| L1' = 1, L1x' = 0, L1y' = 1, a' = a = 90°
| L2' = 0.6, L2x' = 0.6, L2y' = 0, a' = a = 0°
| L3' = 1.1662, L3x' = 0.82,L3y' = 0.82, a' = a = 135°
|
| A-----B
| A | |
| * L3' |
| L1'| * |
| | * |
| D-----C
| L2'
|
To which Marcel Luttgens replied:
| When using cos(a) and sin(a) to calculate the projections of
| the contracted arm, one gets indeed the results that you are
| presenting above, but the "contradiction" is not "huger" than those
| obtained by SR in the case of a rotating disk.
Meaning that Marcel Luttgens admit that my application of his
"transform" with invariant angle is correct.
And the results of the transform are self contradictory.
So what are you saying Marcel?
You admit that your "transform" correctly applied yields
self contradictory results, but you do _not_ admit that
your "transform" is self contradictory? :-)
> But why don't you admit
> that your demonstration of the non-invariance of the angle a
> is logically inconsistent?
It was a rather interesting observation that when you
transformed the square object to "the ether frame", then
you didn't use _your_ transform with invariant angles,
but you used the correct Lorentz transform with non invariant
angles.
Paul B. Andersen wrote:
| Please explain how you think a square 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?
Marcel Luttgens replied:
| 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 !
I does indeed check out.
And the angle of the diagonal a = 45° in the "square frame"
has transformed to the angle a' = 59° in the "ether frame".
Why is it that the transform you insist is inconsistent
yields consistent results, while the transform you say
is correct yields self contradictory reslults?
The answer is rather obvious, no? :-)
Paul
MLuttgens
>Date : Sun, 21 Nov 1999 19:53:02 GMT
>
>::: mlut...@aol.com (MLuttgens)
>::: The arm is contracted by sqrt(1-(v*cos(a))^2), not by
>::: sqrt(1-(v*cos(a'))^2).
>
>:: thr...@sheol.org (Wayne Throop)
>:: That IS what I just said. For about the 37th time
>: But Wayne Throop said in another post
>:: there is no such thing as length contraction by sqrt(1-(v*cos(a))^2)
>
>: mlut...@aol.com (MLuttgens)
>: I find this obviously contradictory, despite any slim, slick and
>: crooked counter argumentation.
>
>They MLuttgens is being dishonest. He knows very well, because I have
>explained in extreme detail in postings he has responded to, exactly
>what I meant by those two statements that you have snipped from context.
>For MLuttgens to pretend to find them contradictory at this point is
>ridiculous, hence I conclude, dishonest.
>
>The term "length contraction" refers to an anisotropic effect in LET.
>It always occurs by a factor of sqrt(1-v^2).
>
>In LET, there is NO "length contraction" which has BOTH a ratio of
>sqrt(1-(v*cos(a))^2) AND preserves the angle a. (0 < a < 90)
>
Nevertheless, the oblique arm is contracted by sqrt(1-(v*cos(a))^2),
where the angle a is preserved. However, according to your
illogical demonstration, its projection is given by
L*sqrt(1-(v*cos(a))^2) * cos(a').
Claiming that the angle a is not preserved relies on a wrong
assumption.
The contraction of the parallel arm by sqrt(1-v^2) does NOT
imply that the projection L*cos(a) of the oblique arm is
contracted by sqrt(1-v^2), and indirectly, that the projection of
the contracted arm of length L'=L*sqrt(1-(v*cos(a))^2) is not
given by L' * cos(a).
In fact, the only factor relating the contraction by sqrt(1-(v*cos(a))^2)
of an oblique arm to that of a parallel or transverse one is cos(a).
Indeed, if the arm is parallel to the velocity vector, a=0°, hence
sqrt(1-(v*cos(a))^2) becomes sqrt(1-v^2), which is the Lorenz
contraction factor, and if the arm is perpendicular to that vector,
a=90°, and the contraction factor is 1, meaning that the arm is not
contracted at all.
>All of which has been discussed to death for the last two months.
>For MLuttgens to claim he didn't know what I meant at this point...
>well, "dishonest" doesn't even half cover it.
>
I understand that the source of your editorial ambiguities is
your incapacity to detect the logical error in your derivation.
So, I apologize for having described your argumentation as being
"slim, slick and crooked".
>: Nature is not illogical, unlike your demonstration of the non-
>: invariance of the angle a:
>
>What about YOUR OWN demonstration of the change in the angle between frames?
>Is it illogical, too? Where do you claim you make your mistake?
>
Which demonstration? Do you mean the paraphrase of your
demonstration?
>: I don't have to admit your false claims.
>
>You don't have to admit true ones, either, as you have demonstrated.
>
>Wayne Throop
Marcel Luttgens
Wayne Throop
:: sqrt(1-(v*cos(a))^2) AND preserves the angle a. (0 < a < 90)
: mlut...@aol.com (MLuttgens)
: Nevertheless, the oblique arm is contracted by sqrt(1-(v*cos(a))^2),
: where the angle a is preserved.
I specifically said "in LET". And in LET, the angle is not invariant.
In MLuttgens' fantasies, the angle may be invariant; but then (as MLuttgens
himself has shown), the invariance of the angle makes the deformation
of solid objects self-contradictory.
And what's the most striking thing about this?
MLuttgens knows it, because he himself actually gave the values for a
set of contracted lengths of a solid object, showing that the angle
cannot be invariant in general.
So. MLuttgens' claim above is a bald-faced lie, one he knows is false,
and (since I conclude he is not deluding himself) the purpose of which
seems to be to mislead other people, pure and simple.
All the rest of MLuttgens verbiage is secondary to this.
As long as he continues to pretend that the angle is invariant when
he knows very well it is not (having shown it himself), then all his
arguments and "proofs" and "demonstrations" or whatever are just
diversions and obfuscation.
Mind you, he'd like you to THINK there's disagreement about something
else, like the length of a diagonal arm, or whatever he can come up
with to divert attention. But underlying his entire current
generation troll is his pretense that the angle is invariant.
If anybody reading doubts the angle differs between coordinates,
see MLuttgens' own demonstration to the contrary, below:
::: Nature is not illogical, unlike your demonstration of the non-
::: invariance of the angle a:
:: What about YOUR OWN demonstration of the change in the angle between
:: frames? Is it illogical, too? Where do you claim you make your mistake?
: Which demonstration? Do you mean the paraphrase of your
: demonstration?
No, I mean MLuttgens' own demonstration in
message <19991110082306...@ngol06.aol.com>
Where he takes the figure
>: L
>: A----------B
>: | |
>: L| |
>: | | -> v
>: | |
>: D----------C
and proceeds to show that
: Let L1=AD=1, and L2=DC=1 So, L3, the hypotenuse AC, is sqrt(2).
: 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
Now, of the arm L3 (that is, AC) MLuttgens shows that
cos(a)| = sin(a)|. Then MLuttgens shows, right here, plain as day, that
while tan(a)=1, tan(a'=(L1/L2) instead, and L2 != L1. Hence a != a'.
In short, MLuttgens has shown quite clearly that the angle cannot
consistently be invariant in solid objects. True, he didn't mention the
obvious consequence of his figures for L1 and L2, but he can't claim he
just overlooked it, since both Paul and I have brought his attention to
this fact several times.
So for some reason, he's pretending not to notice. Shrug.
MLuttgens
>Date : Tue, 23 Nov 1999 02:18:09 GMT
>
>:: In LET, there is NO "length contraction" which has BOTH a ratio of
>:: sqrt(1-(v*cos(a))^2) AND preserves the angle a. (0 < a < 90)
>
>: mlut...@aol.com (MLuttgens)
>: Nevertheless, the oblique arm is contracted by sqrt(1-(v*cos(a))^2),
>: where the angle a is preserved.
>
>I specifically said "in LET". And in LET, the angle is not invariant.
>In MLuttgens' fantasies, the angle may be invariant; but then (as MLuttgens
>himself has shown), the invariance of the angle makes the deformation
>of solid objects self-contradictory.
>
Are you claiming now that the oblique arm is not contracted by
sqrt(1-(v*cos(a))^2) ?
Just look at its x,x' and y'-projections. If you want to get your
L*cos(a)*sqrt(1-v^2) and L*sin(a), you must multiply
the contracted arm's length L'= L*sqrt(1-(v*cos(a))^2) respectively
by cos(a') and sin(a'), but retain the original angle a in L'.
Do you understand your own derivation?
>And what's the most striking thing about this?
>MLuttgens knows it, because he himself actually gave the values for a
>set of contracted lengths of a solid object, showing that the angle
>cannot be invariant in general.
>
I didn't show such thing.
>So. MLuttgens' claim above is a bald-faced lie, one he knows is false,
>and (since I conclude he is not deluding himself) the purpose of which
>seems to be to mislead other people, pure and simple.
>
Either you are trolling, or you don't understand the issue.
>All the rest of MLuttgens verbiage is secondary to this.
>As long as he continues to pretend that the angle is invariant when
>he knows very well it is not (having shown it himself), then all his
>arguments and "proofs" and "demonstrations" or whatever are just
>diversions and obfuscation.
>
You have no serious reason why the angle should vary.
>Mind you, he'd like you to THINK there's disagreement about something
>else, like the length of a diagonal arm, or whatever he can come up
>with to divert attention. But underlying his entire current
>generation troll is his pretense that the angle is invariant.
>
Explain why it should vary.
>If anybody reading doubts the angle differs between coordinates,
>see MLuttgens' own demonstration to the contrary, below:
>
>::: Nature is not illogical, unlike your demonstration of the non-
>::: invariance of the angle a:
>
>:: What about YOUR OWN demonstration of the change in the angle between
>:: frames? Is it illogical, too? Where do you claim you make your mistake?
>
>: Which demonstration? Do you mean the paraphrase of your
>: demonstration?
>
>No, I mean MLuttgens' own demonstration in
>message <19991110082306...@ngol06.aol.com>
>Where he takes the figure
>
>>: L
>>: A----------B
>>: | |
>>: L| |
>>: | | -> v
>>: | |
>>: D----------C
>
>and proceeds to show that
>
>: Let L1=AD=1, and L2=DC=1 So, L3, the hypotenuse AC, is sqrt(2).
>: 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
>
Those values are indeed solutions of the general formula
L'=L*sqrt(1-(v*cos(a))^2), with v=0.8 and the angle a invariant.
>Now, of the arm L3 (that is, AC) MLuttgens shows that
>cos(a)| = sin(a)|. Then MLuttgens shows, right here, plain as day, that
>while tan(a)=1, tan(a'=(L1/L2) instead, and L2 != L1. Hence a != a'.
>
Mathematically, the general formula gives as projections
- for AD (the tranverse arm), L1' = L1*1 = 1
- for DC (the parallel arm), L2' = L2*sqrt(1-v^2) = 0.6
- for AC (the oblique arm), L3' = L3*sqrt(1-(v*cos(a))^2) = 1.1662
The relevant part of the above figure thus becomes
A
|*
| *
L1'=1 | * L3'=1.1662 (from L3=1.4142)
| *
| *
| * <---- C'
D----------C
L2'=0.6
So, the arm AC has contracted to AC'.
Now, Wayne Throop, explain why C' should coincide with C.
If you tell us it is because of some physical contact force, you are
using another theory than LET or SR.
In fact, there is no reasonable explanation why the angle a
should vary.
>In short, MLuttgens has shown quite clearly that the angle cannot
>consistently be invariant in solid objects. True, he didn't mention the
>obvious consequence of his figures for L1 and L2, but he can't claim he
>just overlooked it, since both Paul and I have brought his attention to
>this fact several times.
>
You are speaking now of solid objects. Do you imply that some
force is the cause of the non invariance of the angle a?
>So for some reason, he's pretending not to notice. Shrug.
>
>
>Wayne Throop
Marcel Luttgens
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Mon, 22 Nov 1999 11:05:55 +0100
Mathematically, the general formula gives as projections
- for AD (the tranverse arm), L1' = L1*1 = 1
- for DC (the parallel arm), L2' = L2*sqrt(1-v^2) = 0.6
- for AC (the oblique arm), L3' = L3*sqrt(1-(v*cos(a))^2) = 1.1662
The relevant part of the above figure thus becomes
A
|*
| *
L1'=1 | * L3'=1.1662 (from L3=1.4142)
| *
| *
| * <---- C'
D----------C
L2'=0.6
So, the arm AC has contracted to AC'.
Now, Paul, explain why C' should coincide with C.
If you tell us it is because of some physical contact force, you are
using another theory than LET or SR.
In fact, there is no reasonable explanation why the angle a
should vary.
>
>Paul
Marcel Luttgens
Wayne Throop
: Are you claiming now that the oblique arm is not contracted by
: sqrt(1-(v*cos(a))^2) ?
My WORD but you are a persistent little troll.
You'll say ANYTHING to pretend there's some disagreement
about the length of the arm, won't you.
There is, of course, no disagreement about the length of
the arm, as MLuttgens knows perfectly well by now, since
it has been repeated over and over and over literally dozens of times.
L'=L*sqrt(1-(v*cos(a))^2). Nobody ever said any different.
The real issue, which MLuttgens knows very well, is that I'm
pointing out that a != a'. Thus, the rod's alteration between
the unprimed and primed frame is not a longitudinal contraction.
Again, MLuttgens knows all of this. He's just pretending there's a
controversy, so he can continue to pretend he doesn't realize that the
angle's invariance is logically untenable. Apparently he gets his
jollies this way. As I said, a troll.
:: MLuttgens [] actually gave the values for a
:: set of contracted lengths of a solid object, showing that the angle
:: cannot be invariant in general.
: I didn't show such thing.
Sure he did. See <19991110082306...@ngol06.aol.com>
: Explain why it should vary.
But MLuttgens has already done such a good job of showing
why it must vary. See <19991110082306...@ngol06.aol.com>
: Mathematically, the general formula gives as projections
: - for AD (the tranverse arm), L1' = L1*1 = 1
: - for DC (the parallel arm), L2' = L2*sqrt(1-v^2) = 0.6
: - for AC (the oblique arm), L3' = L3*sqrt(1-(v*cos(a))^2) = 1.1662
: The relevant part of the above figure thus becomes
:
:
: A
: |*
: | *
: L1'=1 | * L3'=1.1662 (from L3=1.4142)
: | *
: | *
: | * <---- C'
: D----------C
:
: L2'=0.6
Here we see diagrammed exactly why MLuttgens' interpretation is inconsistent.
C is not located at C, according to MLuttgens bizarre idea
that the angle is invariant.
It's really quite obvious. L1/L3 is cos(a'), and L2/L3 is sin(a').
And therefore, a cannot possibly be equal to a'.
: Now, Wayne Throop, explain why C' should coincide with C.
Because it's a given of the problem. We are considering three points on
an object. C is a point. A point cannot have two locations (though it
can, of course have any number of coordinates to describe its location). But
what MLuttgens is doing is assigning it two different coordinate
positions IN THE SAME COORDINATE SYSTEM, which of course is so
imbecilic, that even he has to realize it's invalid; hence the
conclusion I've been forced to: he's a troll. He doesn't actually
believe all this nonesense, he just enjoys seeing how many people he can
confuse. It's the only interpretation that fits the facts.
Paul B. Andersen
! :-)
> If you tell us it is because of some physical contact force, you are
> using another theory than LET or SR.
> In fact, there is no reasonable explanation why the angle a
> should vary.
This is too stupid to answer.
You are trolling.
Paul
MLuttgens
>Date : Wed, 24 Nov 1999 19:09:07 GMT
>
>: mlut...@aol.com (MLuttgens)
>: Are you claiming now that the oblique arm is not contracted by
>: sqrt(1-(v*cos(a))^2) ?
>
>My WORD but you are a persistent little troll.
>You'll say ANYTHING to pretend there's some disagreement
>about the length of the arm, won't you.
>
Your claim " In LET, there is NO "length contraction" which has
BOTH a ratio of sqrt(1-(v*cos(a))^2) AND preserves the angle
a. (0 < a < 90)" was ambiguous.
In fact the angle a is preserved in the contraction factor
sqrt(1-(v*cos(a))^2).
You should have better said that a, according to you, is not
preserved in the formula giving the projections of the contracted
arm. Iow, the length L of the arm becomes L'= L*sqrt(1-(v*cos(a))^2),
and the projections of the contracted arm are L' * cos(a') or
L' * sin(a').
As for me, the angle a is invariant, so the projections of the
contracted arm are L' *cos(a) and L' *sin(a).
>There is, of course, no disagreement about the length of
>the arm, as MLuttgens knows perfectly well by now, since
>it has been repeated over and over and over literally dozens of times.
>L'=L*sqrt(1-(v*cos(a))^2). Nobody ever said any different.
>
>The real issue, which MLuttgens knows very well, is that I'm
>pointing out that a != a'. Thus, the rod's alteration between
>the unprimed and primed frame is not a longitudinal contraction.
>
>Again, MLuttgens knows all of this. He's just pretending there's a
>controversy, so he can continue to pretend he doesn't realize that the
>angle's invariance is logically untenable. Apparently he gets his
>jollies this way. As I said, a troll.
>
>:: MLuttgens actually gave the values for a
>:: set of contracted lengths of a solid object, showing that the angle
>:: cannot be invariant in general.
>
>: I didn't show such thing.
>
>Sure he did. See <19991110082306...@ngol06.aol.com>
>
>: Explain why it should vary.
>
>But MLuttgens has already done such a good job of showing
>why it must vary. See <19991110082306...@ngol06.aol.com>
>
>: Mathematically, the general formula gives as projections
>: - for AD (the tranverse arm), L1' = L1*1 = 1
>: - for DC (the parallel arm), L2' = L2*sqrt(1-v^2) = 0.6
>: - for AC (the oblique arm), L3' = L3*sqrt(1-(v*cos(a))^2) = 1.1662
>: The relevant part of the above figure thus becomes
>:
Ooops, the general formula gives as contracted lengths, etc...
>:
>: A
>: |*
>: | *
>: L1'=1 | * L3'=1.1662 (from L3=1.4142)
>: | *
>: | *
>: | * <---- C'
>: D----------C
>:
>: L2'=0.6
>
>Here we see diagrammed exactly why MLuttgens' interpretation is inconsistent.
>C is not located at C, according to MLuttgens bizarre idea
>that the angle is invariant.
>
You surely mean, C' is not located at C. Then, I agree.
>It's really quite obvious. L1/L3 is cos(a'), and L2/L3 is sin(a').
>And therefore, a cannot possibly be equal to a'.
>
No, L1/L3=sin(a), and L2/L3=cos(a).
Remember that L3 is the length of the oblique arm, L1 that
of the transverse arm and L2 that of the parallel arm.
L1, L2 and L3 are the lengths in the interferometer frame.
You probably mean, L1'/L3'=sin(a') and L2'/L3'=cos(a').
>: Now, Wayne Throop, explain why C' should coincide with C.
>
>Because it's a given of the problem. We are considering three
>points on an object. C is a point. A point cannot have two
>locations (though it can, of course have any number of coordinates
>to describe its location). But what MLuttgens is doing is
>assigning it two different coordinate positions IN THE SAME
>COORDINATE SYSTEM, which of course is so
>imbecilic, that even he has to realize it's invalid; hence the
>conclusion I've been forced to: he's a troll. He doesn't actually
>believe all this nonesense, he just enjoys seeing how many
>people he can confuse. It's the only interpretation that fits the facts.
>
No, we are considering three different arms, AD, DC and AC,
with lengths L1=1, L2=1 and L3=sqrt(2)=1.4142.
Due to motion at v=0.8 in the ether, L1 is contracted by the
factor 1, thus L1'=1, L2 by sqrt(1-v^2), thus L2'=0.6 and L3 by
sqrt(1-(v*cos(a))^2), thus L3'=1.1662.
Those arms contract independently from each other, and their
projections are also independent from each other.
The only reason why the angle a should vary is the silly hypothesis
that the arm L3 does not contract by sqrt(1-(v*cos(a))^2), but its
x,x'-projection contracts by sqrt(1-v^2), and its y'-projection
is maintained.
Instead of considering three points on a solid object, hence
obfuscating the issue, try to limit yourself to the consideration
of three independent arms.
>
>Wayne Throop
Marcel Luttgens
Wayne Throop
::: Are you claiming now that the oblique arm is not contracted by
::: sqrt(1-(v*cos(a))^2) ?
:: You'll say ANYTHING to pretend there's some disagreement about the
:: length of the arm, won't you.
: mlut...@aol.com (MLuttgens)
: our claim " In LET, there is NO "length contraction" which has BOTH a
: ratio of sqrt(1-(v*cos(a))^2) AND preserves the angle a. (0 < a <
: 90)" was ambiguous.
You don't really need to go on. People can already see that you'll
go to any lengths whatsoever to pretend there's an ambiguity or
a misunderstanding.
: As for me, the angle a is invariant,
But you yourself showed exactly why the angle can't be invariant.
See <19991110082306...@ngol06.aol.com>,
and consider that L1/L3 and L2/L3 are the sin and cos of a'.
If MLuttgens really wanted to keep pretending the angle is invariant,
he shouldn't have given such a good demonstration exactly why it is not.
: You surely mean, C' is not located at C. Then, I agree.
That th at, (as you well know) is a contraction.
We started with a solid body, with three dots, and three lines
connecting the three dots. You now claim that one dot somehow
became two. You contradict yourself, you KNOW you co ntradict
yourself, but you just aren't honest enough to give it up
and admit you've been trolling. Aparently, you think that if
you just ignore it, maybe you can mislead a few more victims.
: Instead of considering three points on a solid object, hence
: obfuscating the issue,
It is considering three independent arms that obfuscates the issue.
And MLuttgens knows this, since that was the entire point of the
thought experiment; to make it clear your claim that the angle of AC
is invariant conflicts with your claim that objects are contracted
by sqrt(1-(v*cos(a))^2) along their length, with no change in a.
The claim is inconsistent.
If you don't want to consider "points on a solid object", then
consider three long, thin arms, with endpoints at the points of
the triangle ADC. Your claim is that they will "deform" so that
the two endpoints at C are moved apart. This is patently false,
and directly contradicts 1) common sense, and 2) LET (or indeed
any theory involving consistent deformations).
But then. MLuttgens knows all this. He's just trying to
see how many people he can fool, and for how long.
Despicable creature, really.
MLuttgens
<paul.b....@hia.no> wrote :
>> The relevant part of the above figure thus becomes
>>
>>
>> A
>> |*
>> | *
>> L1'=1 | * L3'=1.1662 (from L3=1.4142)
>> | *
>> | *
>> | * <---- C'
>> D----------C
>>
>> L2'=0.6
>>
>> So, the arm AC has contracted to AC'.
>> Now, Paul, explain why C' should coincide with C.
>
> ! :-)
>
>
>> If you tell us it is because of some physical contact force, you are
>> using another theory than LET or SR.
>> In fact, there is no reasonable explanation why the angle a
>> should vary.
>
>This is too stupid to answer.
>
>You are trolling.
>
You can't answer.
>Paul
>
Marcel Luttgens
Paul B. Andersen
| Please explain how you think a square 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
| | |
| 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?
BTW,
This "square object" could be the stone slab on which
the M&M interferometer was mounted. The mirrors were
mounted at the corners, and the "arms" would be the
diagonals A-C and D-B.
Marcel Luttgens' solution:
| > The relevant part of the above figure thus becomes
| >
| >
| > A
| > | *
| > | *
| > L1'=1 | * L3'=1.1662 (from L3=1.4142)
| > | *
| > | *
| > | * <---- C'
| > D----------C
| > L2'=0.6
| >
| > So, the arm AC has contracted to AC'.
| > Now, Paul, explain why C' should coincide with C.
| > If you tell us it is because of some physical contact force, you are
| > using another theory than LET or SR.
| > In fact, there is no reasonable explanation why the angle a
| > should vary.
> Paul B. Andersen responded:
>
> >This is too stupid to answer.
> >
> >You are trolling.
Marcel Luttgens wrote:
>
> You can't answer.
Say, Marcel. Are you for real? :-)
Paul
MLuttgens
>If you don't want to consider "points on a solid object", then
>consider three long, thin arms, with endpoints at the points of
>the triangle ADC. Your claim is that they will "deform" so that
>the two endpoints at C are moved apart.
Yes, this is exactly what I claim.
>This is patently false, and directly contradicts
>1) common sense, and
????
>2) LET
>(or indeed any theory involving consistent
>deformations).
>
I have no ideas about such theory. Do you?
>Wayne Throop
Marcel Luttgens
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Fri, 26 Nov 1999 20:09:15 +0100
So, according to you, the stone slab has contracted in relation
to the angles that the arms of the interferometer make with the
velocity vector.
Are you implying that the slab will contract differently each time
the interferometer is rotated?
Marcel Luttgens
Wayne Throop
:: consider three long, thin arms, with endpoints at the points of the
:: triangle ADC. Your claim is that they will "deform" so that the two
:: endpoints at C are moved apart.
: mlut...@aol.com (MLuttgens)
: Yes, this is exactly what I claim.
: This is patently false, and directly contradicts 1) common sense, and
: ????
Yes. Common sense. That if you bolt the ends of the three arms
to a frame you get a different result th an if you don't. What
force is supposed to move the end of the away from it's initial point.
And yes, common sense. If you suddenly get a different prediction if
you pretend an object has no width at all, from the prediction you get
if you take account of the width of the object, you can be pretty sure
you are fooling yourself with sophisms.
As MLuttgens already knows. But then he's TRYING to fool you with sophisms.
:: 2) LET (or indeed any theory involving consistent deformations).
: I have no ideas about such theory.
Definitely agreed: MLuttgens has no (honest) ideas about LET.
Stephen
mlut...@aol.com (MLuttgens) wrote:
> In article <9435...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :
>
> >If you don't want to consider "points on a solid object", then
> >consider three long, thin arms, with endpoints at the points of
> >the triangle ADC. Your claim is that they will "deform" so that
> >the two endpoints at C are moved apart.
>
> Yes, this is exactly what I claim.
>
> >This is patently false, and directly contradicts
> >1) common sense, and
>
> ????
You tell us how one arm can become two arms in another frame. Use
common sense.
--
"The end of our foundation is knowledge of causes,
and secret motions of things; and the enlarging of the bounds
of human empire, to the effecting of all things possible."
- Francis Bacon, "New Atlantis".
MLuttgens
>Date : Sat, 27 Nov 1999 21:55:39 GMT
>
>:: If you don't want to consider "points on a solid object", then
>:: consider three long, thin arms, with endpoints at the points of the
>:: triangle ADC. Your claim is that they will "deform" so that the two
>:: endpoints at C are moved apart.
>
>: mlut...@aol.com (MLuttgens)
>: Yes, this is exactly what I claim.
>
>: This is patently false, and directly contradicts 1) common sense, and
>
>: ????
>
>Yes. Common sense. That if you bolt the ends of the three arms
>to a frame you get a different result th an if you don't. What
>force is supposed to move the end of the away from it's initial point.
>
And what would happen if the arms' ends are not bolted together?
It is clear that the arms would retain their orientation when
contracting.
>And yes, common sense. If you suddenly get a different prediction if
>you pretend an object has no width at all, from the prediction you get
>if you take account of the width of the object, you can be pretty sure
>you are fooling yourself with sophisms.
>
>As MLuttgens already knows. But then he's TRYING to fool you with sophisms.
>
>:: 2) LET (or indeed any theory involving consistent deformations).
>
>: I have no ideas about such theory.
>
>Definitely agreed: MLuttgens has no (honest) ideas about LET.
>
Who is the sophist? I used the term "theory", referring of course
to a (Throop?) theory involving consistent deformations, not to LET.
>
>Wayne Throop
Marcel Luttgens
MLuttgens
stephe...@deathtospam.hotmail.com (Stephen) écrit :
>Sujet : Re: A simple analysis falsifies SR and LET
>De : stephe...@deathtospam.hotmail.com (Stephen)
>Date : Sat, 27 Nov 1999 23:09:50 +0000
>
>In article <19991127053055...@ngol04.aol.com>,
>mlut...@aol.com (MLuttgens) wrote:
>
>> In article <9435...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :
>>
>> >If you don't want to consider "points on a solid object", then
>> >consider three long, thin arms, with endpoints at the points of
>> >the triangle ADC. Your claim is that they will "deform" so that
>> >the two endpoints at C are moved apart.
>>
>> Yes, this is exactly what I claim.
>>
>> >This is patently false, and directly contradicts
>> >1) common sense, and
>>
>> ????
>
>
> You tell us how one arm can become two arms in another frame. Use
>common sense.
Apparently, you didn't follow the thread.
There are three arms, AD (of length L1=1) perpendicular to
the velocity vector v=0.8, DC (of length L2=1) parallel to v and
AC (of length sqrt(2)/2) making an angle a=45° with DC (or
135° with v).
In the interferometer frame, the three arms are thus the sides of
an isoceles triangle.
Because of the motion of the apparatus in the ether, the arms
will contract according to the general formula
L'=L*sqrt(1-(v*cos(a))^2), hence their lengths become
- for AD (the tranverse arm), L1' = L1*1 = 1
- for DC (the parallel arm), L2' = L2*sqrt(1-v^2) = 0.6
- for AC (the oblique arm), L3' = L3*sqrt(1-(v*cos(a))^2) = 1.1662
and the triangle becomes
A
|*
| *
L1'=1 | * L3'=1.1662 (from L3=1.4142)
| *
| *
| * <---- C'
D----------C
L2'=0.6
So, the arm AC has contracted to AC'.
The question was, why should C' coincide with C?
In this case, the angle a would of course become a', and the
x,x'-projection the contracted oblique arm would be
DC= L2*sqrt(1-v^2) = 0.6, whereas its y'-projection would
be AD=1.
If the arm AC is contracted to AC' without changing its orientation,
its projections would be L3*sqrt(1-(v*cos(a))^2) * cos(a)=
1.1662*cos(45)=0.707 and L3*sqrt(1-(v*cos(a))^2) * sin(a)=
1.1662*sin(45)=0.707.
The difficulty for Wayne Throop and Paul B. Andersen is to
explain the origin of the force that brings about their alleged change
of a to a'. Till now, they don't have found a convincing explanation.
Marcel Luttgens
Stephen
mlut...@aol.com (MLuttgens) wrote:
> Dans l'article <stephenwells-2...@mac008.joh.cam.ac.uk>,
> stephe...@deathtospam.hotmail.com (Stephen) écrit :
>
> >Sujet : Re: A simple analysis falsifies SR and LET
> >De : stephe...@deathtospam.hotmail.com (Stephen)
> >Date : Sat, 27 Nov 1999 23:09:50 +0000
> >
> >In article <19991127053055...@ngol04.aol.com>,
> >mlut...@aol.com (MLuttgens) wrote:
> >
> >> In article <9435...@sheol.org>, thr...@sheol.org (Wayne Throop) wrote :
> >>
> >> >If you don't want to consider "points on a solid object", then
> >> >consider three long, thin arms, with endpoints at the points of
> >> >the triangle ADC. Your claim is that they will "deform" so that
> >> >the two endpoints at C are moved apart.
> >>
> >> Yes, this is exactly what I claim.
> >>
> >> >This is patently false, and directly contradicts
> >> >1) common sense, and
> >>
> >> ????
> >
> >
> > You tell us how one arm can become two arms in another frame. Use
> >common sense.
>
> Apparently, you didn't follow the thread.
I did follow the thread. C' is the end of the oblique arm; if the
transform of C does not coincide with C' then the end of the arm is not
where the end of the arm is, which is dumb. You can complain all you want
but it'll do you no good.
>
>
> The difficulty for Wayne Throop and Paul B. Andersen is to
> explain the origin of the force that brings about their alleged change
> of a to a'. Till now, they don't have found a convincing explanation.
In LET, it's one of the postulates of the theory that arms are
contracted in the direction of motion by a factor gamma. If you're working
in LET, you tell me where the force comes from.
In SR, it's simply a consequence of the geometry that a=/=a'; it's very
easy to work out, and no force is involved.
Paul B. Andersen
>
> In article <383EDADB...@hia.no>, "Paul B. Andersen"
> <paul.b....@hia.no> wrote :
>
> >Date : Fri, 26 Nov 1999 20:09:15 +0100
> >
> >Paul B. Andersen wrote:
> >| Please explain how you think a square 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?
> >
> >BTW,
> > This "square object" could be the stone slab on which
> > the M&M interferometer was mounted. The mirrors were
> > mounted at the corners, and the "arms" would be the
> > diagonals A-C and D-B.
> >
> >Marcel Luttgens' solution:
> >
> >| >
> >| >
> >| > A
> >| > | *
> >| > | *
> >| > L1'=1 | * L3'=1.1662 (from L3=1.4142)
> >| > | *
> >| > | *
> >| > | * <---- C'
> >| > D----------C
> >| > L2'=0.6
> >| >
> >| > So, the arm AC has contracted to AC'.
> >| > If you tell us it is because of some physical contact force, you are
> >| > using another theory than LET or SR.
> >| > In fact, there is no reasonable explanation why the angle a
> >| > should vary.
> >
> >> Paul B. Andersen responded:
> >>
> >> >This is too stupid to answer.
> >> >
> >> >You are trolling.
> >
> >Marcel Luttgens wrote:
> >>
> >> You can't answer.
> >
> >Say, Marcel. Are you for real? :-)
> >
> >Paul
>
> So, according to you, the stone slab has contracted in relation
> to the angles that the arms of the interferometer make with the
> velocity vector.
> Are you implying that the slab will contract differently each time
> the interferometer is rotated?
>
> Marcel Luttgens
Is it still not clear to you how the Lorentz contraction works?
You know the "stone slab", the arm or whatever contracts by
1/gamma in the direction of the motion.
From there, everything else follows.
Which you know pwerfectly well by now.
You demonstated that in your first attempt on the square.
Remember?
Marcel Luttgens wrote:
| 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 !
I does indeed check out.
And the angle of the diagonal a = 45° in the "square frame"
has transformed to the angle a' = 59° in the "ether frame".
What you did above was simply to contract the square by 1/gamma
in the direction of motion. Then the transformation of the diagonal
(which could be the MMX arm), or any other distance on the square
you might wish, follows in the only possible and sensical way.
Paul
Wayne Throop
:: you get a different result th an if you don't. What force is
:: supposed to move the end of the away from it's initial point.
: mlut...@aol.com (MLuttgens)
: And what would happen if the arms' ends are not bolted together? It
: is clear that the arms would retain their orientation when
: contracting.
That was the question I asked YOU, MLuttgens. You claim that
the arms would retain their orientation. Why? How do you justify
such an obviously false claim?
"Arms" are not infinitely thin. Not being infinitely thin, they can
be modeled as a series of square tiles glued together, to any degree
of accuracy you desire. MLuttgens has already said that a solid square
object deforms into a rectangle. Now: do that to every square in the "arm".
What does common sense say will happen?
: Who is the sophist?
I didn't quite say you are a sophist, MLuttgens.
I said you are a liar, and intentionally trying to mislead your readers.
: I used the term "theory", referring of course to a (Throop?) theory
: involving consistent deformations, not to LET.
Case in point. There's no way MLuttgens, at this stage of the
discussion, can honestly claim think LETs "distortions" are either
not distortions at all, or are inconsistent.
MLuttgens
stephe...@deathtospam.hotmail.com (Stephen) wrote :
>Date : Sun, 28 Nov 1999 19:05:12 +0000
I say that the arm AC keeps its original orientation (his angle a)
when contracting to AC'.
How can you seriously claim that "one arm can become two arms
in another frame"?
>>
>>
>> The difficulty for Wayne Throop and Paul B. Andersen is to
>> explain the origin of the force that brings about their alleged change
>> of a to a'. Till now, they don't have found a convincing explanation.
>
>
> In LET, it's one of the postulates of the theory that arms are
>contracted in the direction of motion by a factor gamma.
>If you're working in LET, you tell me where the force comes from.
> In SR, it's simply a consequence of the geometry that a=/=a';
>it's very easy to work out, and no force is involved.
The issue is not contraction, but the alleged rotation of the arm
from a to a'. As for me, geometry cannot be a cause.
In fact, nobody has given a sensible explanation till now.
Marcel Luttgens
Stephen
mlut...@aol.com (MLuttgens) wrote:
> In article <stephenwells-2...@mac016.joh.cam.ac.uk>,
> stephe...@deathtospam.hotmail.com (Stephen) wrote :
>
>
> >> > You tell us how one arm can become two arms in another frame. Use
> >> >common sense.
> >>
> >> Apparently, you didn't follow the thread.
> >
> > I did follow the thread. C' is the end of the oblique arm; if the
> >transform of C does not coincide with C' then the end of the arm is not
> >where the end of the arm is, which is dumb. You can complain all you want
> >but it'll do you no good.
> >
>
> I say that the arm AC keeps its original orientation (his angle a)
> when contracting to AC'.
Which is trivially untrue. We know- because we saw you do it- that you
know what the Lorentz transforms predict; we all know that the x-extent of
the arm is contracted while the y-extent is not; thus we all know that the
angle a' of the arm is not invariant.
If you try to keep a' equal to a then you find that the end of the arm
is in a different place from where it is predicted to be by the consistent
transforms; and since you try to maintain both these incompatible views at
once, I can only conclude that you think the arm becomes two arms.
Why are you so desperate to make a'=a? What's the big deal?
--
Felix qui potuit rerum cognoscere causas - Virgil.
MLuttgens
>Date : Mon, 29 Nov 1999 07:34:13 GMT
>
>:: Common sense. That if you bolt the ends of the three arms to a frame
>:: you get a different result th an if you don't. What force is
>:: supposed to move the end of the away from it's initial point.
>
>: mlut...@aol.com (MLuttgens)
>: And what would happen if the arms' ends are not bolted together? It
>: is clear that the arms would retain their orientation when
>: contracting.
>
>That was the question I asked YOU, MLuttgens. You claim that
>the arms would retain their orientation. Why? How do you justify
>such an obviously false claim?
>
>"Arms" are not infinitely thin. Not being infinitely thin, they can
>be modeled as a series of square tiles glued together, to any degree
>of accuracy you desire. MLuttgens has already said that a solid square
>object deforms into a rectangle. Now: do that to every square in the "arm".
>
>What does common sense say will happen?
>
You come up with arms made up of little square tiles, whereas
Paul B. Andersen is putting forward the stone slab on which the
M&M interferometer was mounted.
Why don't you give the obvious explanation that the rotating Earth
itself is physically contracted, and that its surface and everything
lying on it are accordingly distorted?
[snip]
>Wayne Throop
Marcel Luttgens
MLuttgens
<paul.b....@hia.no> wrote :
>Date : Mon, 29 Nov 1999 08:55:52 +0100
Wayne Throop came up with arms made up of little square tiles,
whereas you are putting forward the stone slab on which the
M&M interferometer was mounted.
Why don't you give the obvious explanation that the rotating Earth
itself is physically contracted, and that its surface and everything
lying on it are accordingly distorted?
Marcel Luttgens
Paul B. Andersen
And your point was?
To divert the attention from your self contradicting "transform"?
Paul
Wayne Throop
: You come up with arms made up of little square tiles
Actually, that's a lie, of course; MLuttgens trying to mislead folks again.
I actually said, you could consider little square-shaped regions of the
arm, and analyzing them, model the deformation of the arm as a whole,
to any degree of accuracy you wish.
Now, what MLuttgens wants you to believe he believes, is that for some
reason, if you just refuse to consider little square-shaped regions
of the arm, for some reason it deforms differently than if you consider
a very long, thing region of the arm having measured angle (a).
But it's fairly clear that even MLuttgens isn't stupid enough to believe
that just accounting for what occurs to each square region of an object is
completely different what occurs to the object as a whole. So he's just
blowing smoke, trying to see how many people he can fool.
He tries to fool as many as he can, by trying to pretend that considering
square regions means "built out of square tiles". But then, he knows just
what he's doing: he's trying to fool people, so he *has* to m isrepresent
the rather clear demonstrations that his claim about the invariance
of the angle is unsupportable either by logic, or any physical theory.
: Why don't you give the obvious explanation that the rotating Earth
: itself is physically contracted, and that its surface and everything
: lying on it are accordingly distorted?
A complete non-sequitur, having nothing to do with the
topic at hand. The topic is whether the angle a equals the angle a'.
But of course, MLuttgens cannot afford to keep to the topic;
the actual subject matter is so simple, MLuttgens can't hide his error
unless he throws out distractions.
MLuttgens
>: Why don't you give the obvious explanation that the rotating Earth
>: itself is physically contracted, and that its surface and everything
>: lying on it are accordingly distorted?
>
>A complete non-sequitur, having nothing to do with the
>topic at hand. The topic is whether the angle a equals the angle a'.
>
I has everything to do with the topic, but you are clearly unable to
go beyond your SR diagrams.
Your state is getting worse.
>
>Wayne Throop
Marcel Luttgens
MLuttgens
<paul.b....@hia.no> wrote :
>> Wayne Throop came up with arms made up of little square tiles,
>> whereas you are putting forward the stone slab on which the
>> M&M interferometer was mounted.
>> Why don't you give the obvious explanation that the rotating Earth
>> itself is physically contracted, and that its surface and everything
>> lying on it are accordingly distorted?
>>
>> Marcel Luttgens
>
>And your point was?
>To divert the attention from your self contradicting "transform"?
>
>Paul
>
Do you imply that the distorsion of the rotating Earth has nothing
to do with the arms' contraction, and correlatively with the
transform?
Marcel Luttgens
Wayne Throop
away from the actual issue at hand, by this ploy:
:: thr...@sheol.org (Wayne Throop)
:: Your claim is that [three arms arranged as a right triangle]
:: will "deform" so that the two endpoints at C are moved apart.
: mlut...@aol.com (MLuttgens)
: Yes, this is exactly what I claim.
At the time, I pointed out that this "directly contradicts 1) common
sense, and 2) LET (or indeed any theory involving consistent
deformations)".
What I neglected to point out is that it is also irrelevant.
Because we weren't discussing a triangle when is then deformed
by setting it in motion in the ether. We were talking about
a single arm, at motion wrt the ether, as measured by two different
sets of calibrated rods and clocks.
So remembering the application, let's strip away a layer of MLuttgens'
diversions, and show why his claim of an invariant angle is not
self-consistent in the original context.
Note: the "arm" we are considering is one "arm" of an interferometer,
and the time for light to traverse the arm is being considered. This
"arm" need not be a physical object. It is defined entirely by two
objects: a light source and a detector, and a mirror. These two objects
are comoving. One way to arrange this is to mount them on the two ends
of an "arm", but the arm is not physically necessary, as MLuttgens
himself has pointed out. The situation is fully defined by the two
objects themselves; the position and orientation of any "arm" is DEFINED
by the positions of these two objects.
So. The line between these two points forms an angle with their common
velocity vector in the ether. If we take comoving measuring rods and
measure the (clearly, constant) separation between the two objects in
the direction of motion as Sx, and the (likewise constant) separation
between the two points along the direction perpendicular to the
direction of motion as Sy, then that angle is given by (Sy/Sx)=tan(a).
If we choose coordinates so that the light source/detector is at the
origin, the mirror is located at (x=Sx,y=Sy); that is, these
measurements-with-comoving-rods are the coordinates of the mirror.
Now. The same mirror, moving in the same way, but considered with
coordinates laid out by measures with rods and clocks motionless in the
ether. The (again, constant) separation between the two objects in
their direction of motion is Sx', and the again, likewise constant)
separation between the two points along the perpendicular direction is
Sy', and the angle a' is given by (Sy'/Sx')=tan(a').
So far, we've just been discussing the meaning of x, y, x', y', a,
and a' [1]. We are given exactly two objects, and we set out to
systematically measure the motion, separation, and orientation of those
objects, and we get values for x,y,a when we use comoving measuring rods,
protractors, and if necessary, clocks. And we get values for x',y',a'
when we use rods, protractors, and clocks at rest in the ether.
In this context, it is MLuttgens' claim that 1) all rods (eg, a
hypothetical rod which happens to exactly span the distance from the
source/detector to the mirror, or for another eg, the rods used to
measure Sx and Sy, will all become shorter by a factor of
sqrt(1-(v*cos(a))^2), and 2) the angle a will equal the angle a'.
MLuttgens' claim is clearly inconsistent. I reinforce the relevant
facts: there is only one source/detector, and only one mirror, and we
are examining them at only one instant of time. In LET[1], the values Sx
and Sy are their measured-with-distorted-rods separation at time t'=0,
measured with comoving rods. The values Sx' and Sy' are that exact same
separation at that exact same time, measured with rods at rest in the
ether. According to MLuttgens' rules, Sy=Sy' and Sx<>Sx', therefore it
is impossible that a=a' as he claims.
It's beyond any rational dispute. MLuttgens' claim is not
self-consistent. The contraction he prescribes for a rod comoving with
source/detector and mirror, which has endpoints always colocated with
them, is inconsistent with the contractions he prescribes for the
measuring rods used to measure Sx, Sy, Sx' and Sy'.
MLuttgens' entire ploy of claiming that the end of the spanning rod is
somehow moved away from the ends of the measuring rods is only an
attempt to divert attention away from the original context, which was
two different measurements of one rod. The measuring rods used to
measure x and y separations by have endpoints at the ends of the
spanning rods, by definition of "to measure".
Insofar as he could divert attention to discussing "distortion" of an
object in two physical situations,MLuttgens could pretend his position was
self-consistent. But when we restore the context he tried to discard,
we can see clearly why his prescription is not self-consistent, and
exactly what wool he was trying to pull over folks' eyes.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
[1] Note that this terminology has grown out of previous discussion,
and is slightly different than usual uses of primed and unprimed
variable names. The x,y,a are "local" values (that is, values
gotten from comoving rods and protractors), and x',y',a' are the
values gotten from rods and protractors at rest in the ether. It's
unclear whether this "just grew" out of the previous discussion, or
whether M Luttgens i ntentionally arranged for this notation to be
deliberately misleading. But intentional or not, don't be misled;
refer back to the definitions above if there's any doubt what is meant.
[2] Roberts takes "LET" to refer to a theory with coordinate systems
using "local" time and "local" distances. For this discussion, I'm
talking about spatial measurements made simultaneously at some
particular LET-absolute-that-is-nonlocal time; that is, the
"comoving coordinates" are using measured distances, but absolute
times. This is adequate for the purposes here, since none of the
discussion at this level involves problems of simultaneity; the
spacelike hyperslice formed by t'=0 is perfectly adequate for both
measurements; it is in those terms that "length contraction" is
defined in LET.