Re: The genius of the Absolute

From: Androcles (dummy_at_dummy.net)
Date: 01/12/05 

Date: Wed, 12 Jan 2005 22:24:27 GMT

"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
news:41E5B1EC.6050708@mail.verizon.net...
>
>
> Androcles wrote:
>
>>"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
>>news:41E592FF.6020100@mail.verizon.net...
>>
>>>Androcles wrote:
>>>
>>>
>>>>"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
>>>>news:41E5848F.6090300@mail.verizon.net...
>>>>
>>>>
>>>>>Androcles wrote:
>>>>>
>>>>>
>>>>>
>>>>>>"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
>>>>>>news:41E57925.1040404@mail.verizon.net...
>>>>>>
>>>>>>
>>>>>>
>>>>>>>Androcles wrote:
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>>>In a reference frame in which the apparatus is stationary, x
>>>>>>>>>and y don't
>>>>>>>>>change as the apparatus is rotated (assuming x and y are the
>>>>>>>>>arm-lengths).
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>Exactly. So you prove nothing. However, Einstein's postulate is
>>>>>>>>"light is always propagated in empty space with a definite
>>>>>>>>velocity
>>>>>>>>c which is independent of the state of motion of the emitting
>>>>>>>>body"
>>>>>>>>Reference :
>>>>>>>>http://www.fourmilab.ch/etexts/einstein/specrel/www/
>>>>>>>>
>>>>>>>>The Earth is moving in the empty space, which is the reference
>>>>>>>>frame.
>>>>>>>>Are you trying to deliberately obfuscate the problem?
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>Empty space doesn't have a particular reference frame in
>>>>>>>Einstein's theory.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>Then light will be source dependent.
>>>>>>Androcles.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>Nope, not according to the laws of electromagnetism.
>>>>>
>>>>>
>>>>Which one? Gauss, Faraday or Ampere?
>>>>Androcles.
>>>>
>>>>
>>>The fact that the speed of light is independent of the source can
>>>only be derived using all of Maxwell's laws.
>>>
>>
>>When you can read a graph, we'll discuss it. Until then, it's a waste
>>of time.
>>Androcles.
>>
>
> I can read your graphs, but with no descriptions of what is being
> measured, I don't know what they mean. For example, on the graph you
> posted at
> http://www.androc1es.pwp.blueyonder.co.uk/New_Folder/00918-ck.gif one
> axis measures time and the other measures "m", which I would guess is
> meters--but what is it that is being measured in meters? Without
> context, a graph like that is meaningless--if you are willing to
> explain the context, I will try to address it.

'm' is magnitude (actually, apparent magnitude). The definition of
magnitude
 http://liftoff.msfc.nasa.gov/academy/universe/MAG.HTML

"When astronomers began to accurately measure the brightness of stars
using instruments, it was found that each magnitude is about 2.5 times
brighter than the next greater magnitude. This means a difference in
magnitudes of 5 units (from magnitude 1 to magnitude 6, for example)
corresponds to a change in brightness of 100 times."

You are seeing a change of 16-9 = 7 magnitudes.
This is AFTER the brightness was first noticed and measurement began.
The curve you see is typical of recurrent novae, although no two are
identical.

So... we are looking at light that has travelled a huge distance.

What is going on, that this curve should be typical?

What do you think would happen if I mounted a machine gun on a carousel
or Ferris wheel, kept it aimed in your direction, just like the seats on
a Ferris
wheel, and proceeded to pepper a steel shield you were hiding behind, a
couple
of miles away. Bullets from the bottom of the wheel would travel the 3
miles
at c+ v.sin(phi) = muzzle velocity + tangential velocity of the wheel,
taking
time (say) 3 miles/(1000 - 5) mph = 0.003015 hours, or 10.85 seconds.
Bullets from the top of the wheel take 3 miles/(1000+5)mph = 10.75
seconds.
Not much difference in the rate of arrirval, is there?
What about from 300 miles away? 1085 seconds and 1075 seconds, right?
Suppose it takes 10 seconds for the wheel to turn the gun from the
bottom
to the top, then the time of arrival of bullets from the top is
1075+ 10 seconds = 1085, exactly the time of arrival of bullets
 from the bottom. So your shield is hit by two bullets at the same
instant,
one from the bottom and one from the top. Actually, you are also hit
by all the bullets in between as well. Now, it is a lot of work
calculating
1,000,000 bullets, I think you'll agree, so that task is best left to a
computer.
Not only that, but the ferris wheel isn't a wheel at all, it is an
ellipse. Much
more calculation, right?
But that's what computers are for, so why not do it right?
And YOU can take advantage of one, because you have one.
All you need is the program. I've even written it for you, and its FREE!
You can specify the parameters of
Distance,
Period
Eccentricity
Semimajor Axis
Yaw
Pitch
Roll
and how many bullets to fire,
and it will plot the arrival of the bullets as a function of time.
Then you can toddle off to the library, find some nice curves
that some kind astronomers have drawn for you but do not
themselves understand the reason for, and recreate those curves
on the computer screen, as I have done.

Now, If Maxwell had had a computer and a program like mine,
he would still have agreed with Gauss and Faraday. But he would
not be saying there was any aether. He would have been saying
the speed of light in the vacuum of space is constant with respect
to the source at the time of emission.
But he didn't have a computer, or a lifetime to calculate 1,000,000
bullets, which would be pretty boring anyway.

>
> In the meantime, regardless of what that graph shows, it doesn't
> change the fact that you have not addressed the issue that Maxwell's
> laws are all based on experiments with electricity and magnetism that
> are extremely well-confirmed, and which don't involve light at all;

I don't need to. I have the data and I have the model. It is up to
Maxwell
(or you) to make his laws fit reality, instead of guessing.
You won't have to change very much, just the error band that all your
extremely well-confirmed experiments have shown to within the limits
of experimental error.

> Once you have these laws, you can show mathematically that together
> they imply the velocity of electromagnetic waves is always
> 1/squareroot(epsilon_0*mu_0), and that this "coincidentally" happens
> to match the observed speed of light very accurately, even though the
> experiments to determine epsilon_0 and mu_0 have nothing to do with
> light.

You are right. "Coincidently". Actually, contrived. What is the speed of
sound in a vacuum? What is the volume of a gas at zero degrees Kelvin?
Just extrapolate, easy enough. Trouble is, there is no sound in a
vacuum.
There are no gases at zero Kelvin. That doesn't matter to you, though.
You believe the vacuum of space has the identical properties of aether,
the equations say so, and it has been experimentally confirmed.
What has been experimentally confirmed, old son, is that the speed
of light in a vacuum is source dependent, billions of times by thousands
of stars.
And YOU are walking around with you head in a book instead of
looking up, telling me I'm wrong, your book says so. I'm looking
at REAL data, EMPIRICAL data, and I don't give a hoot what
your book says, if your "laws" do not agree with the data, Nature
is breaking your laws and she don't care. Now quit telling Nature
what to do, she ain't listening to you, Maxwell, Einstein or anyone
else.

Androcles.

 

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