In sci.physics, Henri Wilson
HW@.
wrote
on Tue, 01 Jul 2003 10:44:42 +1000
:
On Wed, 25 Jun 2003 04:08:31 GMT, The Ghost In The Machine
wrote:

In sci.physics, Henri Wilson
HW@.
wrote
on Tue, 24 Jun 2003 06:39:41 +1000
:
On Mon, 23 Jun 2003 02:01:57 GMT, The Ghost In The Machine
wrote:

In sci.physics, Henri Wilson
HW@.


If B and C are on Earth and A is one pair of a spectroscopic
double, the light speed is still c and it undergoes a
red or blueshift.

At least, such is my understanding.

Are you MAD, Ghost?

Hardly. Theory states that light can cross the Universe
in 0 seconds (subjective). This is one of those weird things
in SR that bends or blows minds; *we*, in our relatively
slow-moving (10^-4 c at the very most) reference frame see
it taking billions of years.

Don't you believe it.

It's not clear exactly how one can measure light degradation
from distant stars, but that would be the most obvious
method by which one could prove it; the position probability
distribution of a particle will degrade over time if that
particle is moving slower than light.

I think there is a lot we don't know.




(Personally, I'm wondering how local matter density affects
lightspeed. This "acceleration constant" appears slightly
spurious.)

Of course.



tAB=L/c+v, tBA=L/c-v

I'm assuming (c+v) and (c-v) for the sake of argument; otherwise
this makes little sense.


total=2Lc/(c^2+v^2)

obviously this should be 2Lc/(c^2-v^2)

or= (2L/c)/(1-(v/c)^2)

Your sign is off; the second term should be (2L/c)/(1+(v/c)^2).

No it shouldn't!
One of mine WAS off but you picked the wrong one.

Oops, you are correct here.

[begins to write on the board 1,000 times:

(a-b) * (a+b) = a^2-b^2
...
(a-b) * (a+b) = a^2-b^2

All done! :-) ]



get it?
Here, OWLS is c.




No, TWSLS (two-way static lightspeed) is assumed c in this case.

Yes, a bit confusing, sorry. 'c' requires some kind of absolute
frame, here.

'c' requires nothing of the sort; it's merely an
arbitrary constant here, or perhaps a light
measurement with respect to a special frame
at rest and spatially distant from a light source.

There you go, "a special frame at rest".
What do you mean by 'rest', Ghost?

A classical notion, that. There is no "special" frame.
Everyone sees light (TWLS) traveling at c.



'c' is OWLS - and also TWLS - in that frame.

'c' is TWLS, if that. No one has ever measured OWLS,
as far as I know.

'c' would also be OWLS when measured by an observer at rest
in an 'absolute aether'.
c is never TWLS in true relativity. It only appears so
because of Einstein's definition.

Um...what is "Einstein's definition"? Refresh my memory here.



v is the observer's velocity wrt that frame.

'v' is the relative frame velocities. The observer cannot
be in the special frame as he's moving with respect thereto.

Of course he can. If one assumes an absolute frame (which I don't)
then anything can move wrt that frame.
What are you trying to say?

Good question. :-) But I probably should have included something
such as "with respect to the light source". Or perhaps to
"the other frame".




OWLS is either c+v or c-v in your equations, depending on
direction. TWLS (or, if you prefer, TWDLS) is c/(1+(v/c)^2).

Yes OK.
Note: should be c/(1-(v/c)^2).

Right.



In any event, you need to do a similar calculation with velocity
at *right angles* to the lightspeed, as well. This is easily
enough done; recall that the light is traveling along the
hypotenuse of a triangle; side of light travel is L, side of
destination travel is vt'AB, where t'AB is the time it takes for
the light to get there and v is the "crosswind" velocity. The
light traverses the hypotenuse of a right triangle in accordance
with this equation:

t'AB = sqrt(L^2 + v^2t'AB^2) / c

We need to solve for t'AB. Squaring:

t'AB^2 = (L^2 + v^2t'AB^2 /)/c^2

(note that t'AB is positive so there's no problems here with
introducing extraneous roots later on).

Multiplying both sides by c^2:

c^2tAB^2 = c^2L^2 + v^2tAB^2

Collecting terms:

t'AB = sqrt(c^2L^2 / (c^2 - v^2))
= L / sqrt(1 - v^2/c^2)

Intuitively this result makes sense (at least from a Newtonian
standpoint) as it takes longer/more effort to fly, walk or
proceed in a crosswind or crosscurrent than it does in a
still medium.

This is the aether view, yes.

The classical luminiferous aether view, yes, not the Haether view,
which admittedly I'm still trying to analyze.

My Haether theory is still evolving but is looking better every day.




If we assume L = 10m (suggested by

http://micro.magnet.fsu.edu/primer/l...edoflight.html

in its description of the MMX experiment) and
v = 30 km/s = 10^-4c (the approximate orbital speed
of dear old mother Earth), we get

tAB - t'AB = L/(c - v) - L/sqrt(c^2 - v^2)
= about 3.33 picoseconds, or 1 mm.

If MMX used green light (500 nm) this would be very readily visible,
but hard to measure for various reasons -- mostly because it's
hard to shield light from the hypothesized luminiferous aether.

A more involved computation emulating two measurements of
TWLS, conducted simultaneously (one at right angles, one
"head-on"):

tAB + tBA - t'AB - t'BA = L/(c-v) + L/(c+v) - 2*L/sqrt(c^2 - v^2)

suggests a far smaller deviation -- about 3.33 femtoseconds,
or 2 wavelengths of 500 nm light. Still readily detectable
with the MMX apparatus, though, since it was designed to
rotate freely on a mercury-filled trough with a wooden float.

Except that MMX did not detect anything. Hence your assumption
of local lightspeed invariance. (The aether in this case
is held into place by matter!)

Ghost, we are well aware of the MMX analysis. Some of us
know that the null result was due to the fact that the
theory behind the experiment was faulty.

The theory *was* faulty. An absolute luminiferous aether was
nicely disproven by that experiment.

How can you prove that something doesn't exist?

By showing that the theoretical effects don't occur, for
the most part. Admittedly, MMX can't distinguish between
nonexistent luminiferous aether and lightspeed-source-local-invariance.

There may still be an aether, but it's now a fluid thing.

You must be able to test a hypothetical property of that something.
If it doesn't exist it doesn't have any testable properties.

We don't know it exists until we give it properties. :-)
A propertyless entity is a bit like the empty set: there's
exactly one empty set.



The theory still *is* faulty, in light of the "acceleration of
galactic recession". A galaxy, AFAIK, is a bunch of stars
all radiating in different directions with a central massive
black hole. Unless one assumes that the black hole is somehow
outfitted with a space drive of some sort there's no method
by which it can accelerate. I suspect a computation error;
the most logical one might have the permittivity and
permeability of free space being affected by local matter
density somehow. At least, such is my naive view on the
matter.

It's clear lightspeed is affected by matter -- light in
glass is slower than light in vacuum. Space is not
a vacuum (although it's damned close).

It is also filled with turbulent Haether of variable 'density'.

Interesting. Not sure how the "density" would affect
lightspeed.




What you have tries to do is define what is actually
meant by OWLS and TWLS.

Well, *somebody* has to do it. :-) OWLS has the problem that
it needs to be measured by two synchronized clocks at
different positions. A precise definition is very important.

I think I have found a way to measure OWLS with only one clock.
More about that later.

I'd be interested in the details of that, and so would a number
of others I suspect.

[snip for brevity]

Now, since this is crossposted to sci.astro, one can also
note that many have observed spectroscopic binaries.
These binaries would have slightly different observed
phenomena, especially eclipsing binaries at some distance,
were the light quanta being shoved at us at different
speeds because of the movements of the two stars.

Apparently this was first postulated by a Walter Ritz,
later disproved by Willem de Sitter. Both are mentioned in

http://www.asa3.org/ASA/topics/Astro...6Phillips.html

Ritz asssumed source dependency.

Exactly. His theory was shot down.

Not necessarily.
Not entirely.


[snip for brevity]

See my animation www.users.bigpond.com/hewn/photons.exe
(a very small file download)
It poses a big unanswered question.

Yeah: how to run it on a Linux system. :-)

Surely you know someone with windows. The demo takes
only seconds to download and run.

You really need to get into Java, sir. :-) Then you and I
won't have these technical glitches.

[.sigsnip]

--
#191,
It's still legal to go .sigless.