GETTING RID OF EINSTEIN RELATIVITY

On Fri, 14 Sep 2007 02:47:02 +0000 (UTC), bz
wrote:

HW@....(Henri Wilson) wrote in
news
On Wed, 12 Sep 2007 15:35:05 +0000 (UTC), bz
wrote:

bz wrote in
9.198.139:

If the source changes motion or even ceases to exist after the light
has been emitted, the light neither knows nor cares.


In fact, as the 'source' is most often an ion, atom or molecule in an
excited state,

and the act of emission makes the source 'cease to exist' [in the
excited state]

then there is no longer a 'reference' to which to tie the light, once it
has been emitted,

so there is no way for the light to know what it's 'mother' does, once
it has left home.

The claim is that a photon initially leaves its source at c relative to
the source frame, at the instant of emission.

That is NOT Sean's claim. He claims that the light must continue to
maintain a velocity of c wrt the source[not the sources iFoR] AFTER
emission, no matter how the source moves.

That frame doesn't change just because the source goes somewhere
else....

Can't you undertand that?

I understand that, Sean doesn't.


WHY it leaves its source at c is something that physicists should be and
would be investigating if Einsteiniana hadn't raised its ugly head..

Einstein said that it DOES leave the point of emission [as measured in an
iFoR co-moving with the source at the instant of emission] at c.

He got that one right...

Einstein also said that the exact same photon moves at c as measured in ANY
iFoR.

He got that one wrong....stolen directly from LET.

Henri, I am surprised at you. You should know by now that science doesn't
investigate 'why' questions.

That shows how much YOU know about science.

WHY DOES THUNDER FOLLOW A LIGHTNING STRIKE, BOB?

Do you still believe it shows the gods are angry?

Haven't you learned that yet? Religion and philosophy ask and answer the
'why' questions.

Science asks and attempts to answer 'how' things behave questions.

Oh crap!....what would you know about it anyway?


www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

"Henri Wilson" HW@.... wrote in message
...
On Fri, 14 Sep 2007 03:59:50 -0700, George Dishman

wrote:


Henri Wilson wrote:
On Wed, 12 Sep 2007 23:07:11 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
news On Fri, 31 Aug 2007 10:14:38 +0100, "George Dishman"
wrote:
...

George, my simulations are accurate. They also match most star
curves.

George, you admit that the SR sagnac diagram has the rays closing on
the
source at c+/-v, when viewed in the rest frame.

Changing the subject again Henry? We were
talking about the ballistic version.

Do you not then agree that to a third orbserver, the light from a
distant
orbiting star would 'close on' another star at c+v.sin(at)?

No, it would close at c is SR.

You got that wrong.

I got it right, you may be thinking of ballistic theory
instead of SR.

Most of your colleagues use the term 'closing speed' in this contaxt.

I think the term originally came from the military and
referred to the speed at which an interceptor 'closed'
on a target.

It does
not have to equal c.

No, but above the light would close at c on the third
star which is what you asked. It would move away from
the first two at some other value and it would close
on the third at a variable speed in ballistic theory
but not in SR which was the context of the question.

I think you just made a mistake in your question, we
have never disagreed about any of this.

TWLS is dead constant for the simple reason that light is ballistic.
That is, it moves at c wrt its source and everything at rest wrt the
source.
So naturally tAB=tBA in any TWLS experiment and in that case, TWLS
also =
OWLS
= c

Irrelevant, Sagnac measures the one-way speed.

the fact that you don't understand doesn't make it irrelevant.

You said ".. in any TWLS experiment .." but Sagnac
isn't a two-way test. What you say is true for MMX for
example which _is_ a two way measurement, it measures
th sum of the times, not the difference.

Sagnac measures rotation.

Yes.

It does that by exploiting the phenomenon that light does not reflect from
a
moving mirror at the incident angle and speed.

It does so because the speed is c in the inertial frame
(ingoring refractive index effects) whatever the cause.
Ballistic theory also says the speed after reflection
will be c incidentally.

...but that's not what your sagnac diagram shows...

You need to be clear about which diagram you
mean, I have done both SR and ballistic. Both
are shown correctly.

I am refering to the standard sagnac ring diagram.

No, you said ". that's not what YOUR sagnac diagram
shows .." (my emphasis).

'YOUR' here means 'relativist'

No, it means "George's", you just need to
say which one because I have done several.

The rays are shown to move
at c+/-v wrt the source.
Don't deny it George.

Sorry Henry, I have corrected that error so many
times, I'm not going to repeat it. You know you
are wrong but if you insist on wishing to be
ignorant, that's your choice.

George, your 'correction' is not a correction at all.

Yes it is Henry, you need to learn the terminology
of the field.

The plain fact is, the times are calculated to be d/c+v and
d/c-v....distance
traveled over light speed.

Wrong again, that is distance over the difference
between the light speed and the target speed, or
as we discussed above the distance to be closed
divided by the rate at which that distance
decreases hence "closing speed".

Andersen has written this mistake many times for the world to see.

Paul is correct, you are wrong. You should learn
what he has been trying to teach you.

there are lots of stars out there George...and only a limited number of
misled
astronomers.

Only a fraction are eclising binaries and for those
that have had their luminosity curves studied, you
will find most have also had their spectra recorded
to the point of confirming that they are spectroscpic
binaries.

That's right...and most eclipsing spectroscopic binaries are close and
have
short periods. That means their EM spheres overlap and little or no
brightness
variation is expected..only that due to VDoppler.

Yes, there is only VDoppler in every case studies, you
can offer not a single case where ADoppler can be
_proven_ to exist.

Paul asked me many times to explain why HD80715 isn't a variable. I have
done
that.

You haven't explained it at all Henry, in science
explaining means showing mathematically from the
equation which constitutes your theory. Until you
publish that, you have no explanation.

You have no idea how to use it then, it
should reduce the calculation time by a
factor of about 100,000 from the figures
you gave me.

It doesn't, It is only about twice as fast.

Then you haven't made best use of it. It sounds as
though you are still iterrating along the light path.

No, your 'doppler' method requires fewer sample points but many more
subsequent
calculations at each one.

No, it should allow the same number of samples
round the orbit, whihc is just however many
points you want on your graphs, and then you
do a single set of calculations for each point,
all of which are analytical. You should not have
any iteration along the light path and no
iterative methods used for the Keplerian orbit
if you use the mean anomaly as you independent
variable rather than time.

On emy computer, each set of curves is produced in less that a second with
either method....so it doesn't matter which one I use. It's nice to have
agreement between the two....just for checking the methods....

That's cool then. With my method it should go down
to a few tens of milliseconds which is good enough
for a live version, hence the style of my GUI.


You snipped the equations! Here they are again:

However D/(c+v) + D/(c-v) = 2Dc/(c^2-v^2).....so the time taken is
always
greater than D/c.

Nope, without the planet it is

2 * D/(c-v1)

where v1 is due to direct sunlight. With the planet
it becomes

D/c + D/(c-v1+v2)

where v2 is due to the reflected light. It is obvious
even qualitatively that both effects cause it to
arrive earlier and that is on top of the ballistic
effect which also makes it earlier. Forget it Henry,
it is a non-solution.

Either you don't understand or you are having difficulty in conveying
what you
mean....

Probably I didn't understand because you had trouble
conveying what you meant. Try defining what D and v
are in your equation, I may have guessed incorrectly.

A---------------------S---------------------B

If an object falls under gravity from A towards the star and then on onto
B,
Its final speed will equal that at A.

Yes, but that's a trivial part of the setup. We
are comparing two situations:


S--p---q-----------E
\ \
\ \
\ \
\ \
\ \
Sun

According to your suggestion, the signal from
satellite S is slowed by radiation pressure
from the sunlight all along the path. Consider
in particular parts p and q. Now introduce
Jupiter:

S------q-----------E
/ \
J \
\ \
\ \
\ \
Sun

At point p, Jupiter shades the signal so it is
not slowed by the sunlight, it arrives earlier.
At point q it is still slowed by the sunlight
as before (so no cnhnage from that) but the
reflected light acts in the opposite direction
pushing the signal towards Earth and again
making it arrive earlier.

I took D in your equation as the length of the
shadowed path at p and since you only used one
distance guessed you were assuming the length
at q over which reflected light was significant
was the same (not unreasonable). Your speed v
would be the difference between the speed without
the planet and the speed when it was present. The
equations were still too simple but that's the
only interpretation that made any sense.

The time taken is calculated by integrating dD/v(t).
Any force that efectively reduces the garvity pull will result in an
increased
travel time.

Sure, but when you remove the effect that
counteracts gravity, e.g. by shadowing the
sunlight, then the speed increases and the
signal arrives earlier.

Geoge, all my curves are ADoppler. I have no trouble in matching
observed ones.

Sure, but all your curves are matched to the temperature
and radius variation, not any ADoppler part so they are
universally wrong. Even once you correct that error, it
still means nothing, what you have to do to produce
a proof is show that it _is_ ADoppler, not just that it
_might_ be.

The temperature and radius variations are largely mythical.

Sorry Henry, they are accurately measured according
to ballistic theory and until you take them into
account your curves are meaningless.

George

Henri Wilson skrev:
On Wed, 12 Sep 2007 12:10:25 +0200, "Paul B. Andersen"
wrote:

Pentcho Valev wrote:
The emission theory gives the equations c'=c-v and c'=c+v whereas
special relativity gives c'=c. Which equations: c'=c-v and c'=c+v or
c'=c, are relevant in the interpretation of the Sagnac experiment?

Pentcho Valev
The Sagnac experiment:
- Given an inertial frame which is the reference
for all speeds mentioned below.
That is, all speeds are relative to this non-rotating frame.
- Given a stationary circle with radius r.
- Given a light source moving at the speed v around the circle.
- Assume the light is moving around the circle (infinite number of mirrors).
- Let tf be the time the light emittet in the forward direction
uses to catch up with the source.
- Let tb be the time the light emittet in the backward direction
uses to meet the source.

Prediction according to SR:
---------------------------
The speed of the light emitted in the forward direction is c.
The speed of the light emitted in the backward direction is c.

wrt what?

So we have:
2*pi*r + tf*v = tf*c
tf = 2*pi*r/(c-v)

2*pi*r - tb*v = tb*c
tb = 2*pi*r/(c+v)

delta_t = tf - tb = 4*pi*r*v/(c^2 - v^2)

You have already proved SR wrong. You have calculated travel TIME as d/c+v.

SIC!

[snip]

Paul

Androcles skrev:
"The Ghost In The Machine" wrote in message
...
: In sci.physics.relativity, Jeckyl
:
: wrote
: on Wed, 12 Sep 2007 21:20:11 +1000
: :
: "Androcles" wrote in message
: .uk...
:
: Prediction according to SR:
: The rays do not meet.
:
: Nonsense .. go learn some physics instead of posting crap
:
:
: Bear in mind he's been here since 1999... :-)

If tb doesn't equal tf then the rays don't meet. shrug

If tb doesn't equal tf then there is a phase difference
between the continuous waves when they meet.
That's why the fringes shift.

Actually the clown Andersen lifted his calculation straight
from http://www.mathpages.com/rr/s2-07/2-07.htm

This calculation is so simple that any clown can do it by himself.

Even the clown Androcles can correctly calculate what
the emission theory predicts for the Sagnac experiment:

and it should be

Stationary frame:
tf = (2*pi +alpha)*r/(c+v)

tb = (2*pi - alpha)*r/(c-v)
tb = tf

Rotating frame:
tf = tb = 2*pi*r/c

Well done, Androcles.
The emission theory predicts tf = tb, no phase difference,
no fringe shifts.

Glad to see you know that the Sagnac experiment falsifies the emission theory.

but the idiot left out alpha, clearly shown in the diagram.

But you and I know better, Androcles.
Since you were able to solve your equations, you know that alpha*r = t*v,
so I wouldn't for a minute suspect that you didn't notice the terms
tf*v and tb*v in my equations below.

So who is the idiot that forgot it?

As opposed to mentioned unnamed idiot, WE, you and I, can both correctly
calculate what the emission theory predicts for the Sagnac experiment:

My calculation:

| Prediction according to the emission theory:
| --------------------------------------------
| The speed of the light emitted in the forward direction is c+v.
| The speed of the light emitted in the backward direction is c-v.
|
| So we have:
| 2*pi*r + tf*v = tf*(c+v)
| tf = 2*pi*r/c
|
| 2*pi*r - tb*v = tb*(c-v)
| tb = 2*pi*r/c
|
| delta_t = tf - tb = 0
|
| So emission theory predicts delta_t = 0, while enumerable practical
| experiments shows delta_t = 4Aw/c^2
|
| Prediction wrong - emission theory falsified.

Don't you think you and I have settled the matter once for all now, Androcles?

Paul, in perfect agreement with Androcles

Henri Wilson skrev:
On Fri, 14 Sep 2007 16:23:32 +0200, "Paul B. Andersen"
wrote:
Sagnac falsifies emission theory.
No question about it.

Paul, your own maths show clearly that Sagnac disproves SR. The two travel
times YOU calculated as (distance/light speed) are (D+d)/(c+v) and (D-d)/(c-v).

| Prediction according to SR:
| ---------------------------
| The speed of the light emitted in the forward direction is c.
| The speed of the light emitted in the backward direction is c.
|
| So we have:
| 2*pi*r + tf*v = tf*c

D + d = tf*c

tf = (D + d)/?

I know. Too hard for you.

| 2*pi*r - tb*v = tb*c

D - d = tb*c

tb = (D - d)/?

I know. Too hard for you.



What does that say about light always traveling at c?

It says that this simple calculation is beyond your abilities.
But we knew that.
Further demonstrations unnecessary.
You can stop now.

Paul

"Paul B. Andersen" wrote in message
...
: Henri Wilson skrev:
: On Fri, 14 Sep 2007 16:23:32 +0200, "Paul B. Andersen"
: wrote:
: Sagnac falsifies emission theory.
: No question about it.
:
: Paul, your own maths show clearly that Sagnac disproves SR. The two
travel
: times YOU calculated as (distance/light speed) are (D+d)/(c+v) and
(D-d)/(c-v).
:
: | Prediction according to SR:
: | ---------------------------
: | The speed of the light emitted in the forward direction is c.

But the emitter is moving at v.
Too hard, even for a troll like you.
--


'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein

http://www.androcles01.pwp.blueyonde...rt/tAB=tBA.gif

"Neither [frame] is stationary, which is your problem." -- Blind
"I'm not a troll" Poe.
Ref: ups.com



'we establish by definition that the "time" required by
light to travel from A to B doesn't equal the "time" it requires
to travel from B to A in the stationary system, obviously.' --
Heretic Jan Bielawski, assistant light-bulb changer.

Ref: ups.com


"SR is GR with G=0." -- Uncle Stooopid.

The Uncle Stooopid doctrine:
http://sound.westhost.com/counterfeit.jpg

"What can be asserted without evidence can also be dismissed without
evidence." -- Uncle Stooopid.


"Counterfactual assumptions yield nonsense.
If such a thing were actually observed, reliably and reproducibly, then
relativity would immediately need a major overhaul if not a complete
replacement." -- Humpty Roberts.

Rabbi Albert Einstein in 1895 failed an examination that would
have allowed him to study for a diploma as an electrical engineer
at the Eidgenössische Technische Hochschule in Zurich
(couldn't even pass the SATs).

According to Phuckwit Duck it was geography and history that Einstein
failed on, as if Eidgenössische Technische Hochschule would give a
damn. That tells you the lengths these lying *******s will go to to
protect their tin god, but its always a laugh when they slip up.
Trolls, the lot of them.

"This is PHYSICS, not math or logic, and "proof" is completely
irrelevant." -- Humpty Roberts.

Well, was Einstein's Theory wrong? Will they take away his Nobel Prize?
Inquiring minds want to know. ~Lainie~

~*Lainie~*The StarGazer*~


My Astronomy Website:
http://community.webtv.net/LAINIE121/doc

wrote in message
...
: Well, was Einstein's Theory wrong?

Yes.

: Will they take away his Nobel Prize?

No, his prize was for the photo-electric effect, not his
crackpot theories.


: Inquiring minds want to know. ~Lainie~

Now you know.


: ~*Lainie~*The StarGazer*~
:
:
: My Astronomy Website:
: http://community.webtv.net/LAINIE121/doc

wrote in message
...
Well, was Einstein's Theory wrong?

No.

Will they take away his Nobel Prize?

He received it for his work on the photoelectric
effect.

Inquiring minds want to know.

Now you do.

George

On Sun, 16 Sep 2007 11:46:21 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Fri, 14 Sep 2007 03:59:50 -0700, George Dishman


I got it right, you may be thinking of ballistic theory
instead of SR.

Most of your colleagues use the term 'closing speed' in this contaxt.

I think the term originally came from the military and
referred to the speed at which an interceptor 'closed'
on a target.

It does
not have to equal c.

No, but above the light would close at c on the third
star which is what you asked. It would move away from
the first two at some other value and it would close
on the third at a variable speed in ballistic theory
but not in SR which was the context of the question.

....in SR as well..


You said ".. in any TWLS experiment .." but Sagnac
isn't a two-way test. What you say is true for MMX for
example which _is_ a two way measurement, it measures
th sum of the times, not the difference.

Sagnac measures rotation.

Yes.

It does that by exploiting the phenomenon that light does not reflect from
a
moving mirror at the incident angle and speed.

It does so because the speed is c in the inertial frame
(ingoring refractive index effects) whatever the cause.
Ballistic theory also says the speed after reflection
will be c incidentally.


Then why are the rays shown to be moving at c+/-v wrt the inertial frame of the
source at the moment of emission?

Sorry Henry, I have corrected that error so many
times, I'm not going to repeat it. You know you
are wrong but if you insist on wishing to be
ignorant, that's your choice.

George, your 'correction' is not a correction at all.

Yes it is Henry, you need to learn the terminology
of the field.

The plain fact is, the times are calculated to be d/c+v and
d/c-v....distance
traveled over light speed.

Wrong again, that is distance over the difference
between the light speed and the target speed, or
as we discussed above the distance to be closed
divided by the rate at which that distance
decreases hence "closing speed".

Andersen has written this mistake many times for the world to see.

Paul is correct, you are wrong. You should learn
what he has been trying to teach you.

TIME = (DISTANCE TRAVELED)/(RELATIVE LIGHT SPEED).
OR, as Paul stated so clearly: D/(c+v)


That's right...and most eclipsing spectroscopic binaries are close and
have
short periods. That means their EM spheres overlap and little or no
brightness
variation is expected..only that due to VDoppler.

Yes, there is only VDoppler in every case studies, you
can offer not a single case where ADoppler can be
_proven_ to exist.

George, we often hear the term 'psychological deafness', You suffer from
'psychological blindness'.

Paul asked me many times to explain why HD80715 isn't a variable. I have
done
that.

You haven't explained it at all Henry, in science
explaining means showing mathematically from the
equation which constitutes your theory. Until you
publish that, you have no explanation.

The maths is trivial.
Luminosity variation = VDoppler (almost negligible).

You have no idea how to use it then, it
should reduce the calculation time by a
factor of about 100,000 from the figures
you gave me.

It doesn't, It is only about twice as fast.

Then you haven't made best use of it. It sounds as
though you are still iterrating along the light path.

No, your 'doppler' method requires fewer sample points but many more
subsequent
calculations at each one.

No, it should allow the same number of samples
round the orbit, whihc is just however many
points you want on your graphs, and then you
do a single set of calculations for each point,
all of which are analytical. You should not have
any iteration along the light path and no
iterative methods used for the Keplerian orbit
if you use the mean anomaly as you independent
variable rather than time.

That's not a good method.

Since my curves are 600 pixels in length, I only need 600 sample pairs of
points around the orbit.

On my computer, each set of curves is produced in less that a second with
either method....so it doesn't matter which one I use. It's nice to have
agreement between the two....just for checking the methods....

That's cool then. With my method it should go down
to a few tens of milliseconds which is good enough
for a live version, hence the style of my GUI.

You do it then George. There is nothing wrong with my GUI. It is extemely
simple now...considering thejre are 13 variables and several individual
programs.

You snipped the equations! Here they are again:

However D/(c+v) + D/(c-v) = 2Dc/(c^2-v^2).....so the time taken is
always
greater than D/c.

Nope, without the planet it is

2 * D/(c-v1)

where v1 is due to direct sunlight. With the planet
it becomes

D/c + D/(c-v1+v2)

where v2 is due to the reflected light. It is obvious
even qualitatively that both effects cause it to
arrive earlier and that is on top of the ballistic
effect which also makes it earlier. Forget it Henry,
it is a non-solution.

Either you don't understand or you are having difficulty in conveying
what you
mean....

Probably I didn't understand because you had trouble
conveying what you meant. Try defining what D and v
are in your equation, I may have guessed incorrectly.

A---------------------S---------------------B

If an object falls under gravity from A towards the star and then on onto
B,
Its final speed will equal that at A.

Yes, but that's a trivial part of the setup. We
are comparing two situations:


S--p---q-----------E
\ \
\ \
\ \
\ \
\ \
Sun

According to your suggestion, the signal from
satellite S is slowed by radiation pressure
from the sunlight all along the path. Consider
in particular parts p and q. Now introduce
Jupiter:

S------q-----------E
/ \
J \
\ \
\ \
\ \
Sun

At point p, Jupiter shades the signal so it is
not slowed by the sunlight, it arrives earlier.
At point q it is still slowed by the sunlight
as before (so no cnhnage from that) but the
reflected light acts in the opposite direction
pushing the signal towards Earth and again
making it arrive earlier.

I assume the 'solar wind' is more than just light radiation.

I took D in your equation as the length of the
shadowed path at p and since you only used one
distance guessed you were assuming the length
at q over which reflected light was significant
was the same (not unreasonable). Your speed v
would be the difference between the speed without
the planet and the speed when it was present. The
equations were still too simple but that's the
only interpretation that made any sense.

The time taken is calculated by integrating dD/v(t).
Any force that efectively reduces the garvity pull will result in an
increased
travel time.

Sure, but when you remove the effect that
counteracts gravity, e.g. by shadowing the
sunlight, then the speed increases and the
signal arrives earlier.

That might be an experiment worth trying.

Geoge, all my curves are ADoppler. I have no trouble in matching
observed ones.

Sure, but all your curves are matched to the temperature
and radius variation, not any ADoppler part so they are
universally wrong. Even once you correct that error, it
still means nothing, what you have to do to produce
a proof is show that it _is_ ADoppler, not just that it
_might_ be.

The temperature and radius variations are largely mythical.

Sorry Henry, they are accurately measured according
to ballistic theory and until you take them into
account your curves are meaningless.

George, is our sun changing temperature significantly?
It would appear to be a variable to a distant observer.




www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.