Re: 1c+1c Closing Velocity of Light and Matter

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 01/18/05 

Date: Tue, 18 Jan 2005 06:01:38 GMT

In sci.physics, H@..(Henri Wilson)
<H@>
 wrote
on Tue, 18 Jan 2005 01:59:35 GMT
<r6rou0p7b9hg43qr49khtpd2nhnsu3v2u5@4ax.com>:
> On Mon, 17 Jan 2005 23:08:29 +0100, "Paul B. Andersen"
> <paul.b.andersen@deletethishia.no> wrote:
>
>>The Ghost In The Machine wrote:
>>> In sci.physics, Paul B. Andersen
>>> <paul.b.andersen@deletethishia.no>
>>> wrote
>>> on Mon, 17 Jan 2005 13:48:38 +0100
>>> <csgc77$snm$1@dolly.uninett.no>:
>>>>I hardly think NASA will spend hundreds of millions $
>>>>to test a theory which is thoroughly falsified. :-)
>>>
>>>
>>>
>>> Obviously it's not been thoroughly falsified as some people
>>> still believe in it.
>>
>>It is thoroughly falsified.
>>
>>> Then again, there are those who believe that we faked the
>>> Moon landing, so it takes all kinds. :-)
>>
>>The Earth is flat.
>>Holocost never happened.
>>There is an ancient city on Mars - Cydonia.
>>The speed of light depend on the velocity of the source.
>>I could go on ..
>>
>>Are you saying that none of these are thorougly
>>falsified as some people still belive in them? :-)

I'm not sure, though obviously for those people who believe
in them, they're not falsified. Of course one has to
wonder how seriously to take such individuals...

I'll note for the record that I see a lot of UFOs.
Living near a busy airport does that to a guy... :-)

>
> SR is thoroughly falsified yet people still believe IT!
> "A vertical light beam does not heel over and move diagonally
> at c in the frame of a moving observer".

Doesn't it? Lessee. Time to break out the vector math.
This is for a lightbeam going in an arbitrary direction w,
from an arbitrary starting point K. I assume the usual
two observers and an abstract Euclidean space.

P = wt + K

Of course the beam travels at lightspeed, so

c = ||w||

For the other observer I will use the following notations:

P' = w't'
c' = ||w'||
where P' and t' are defined using the Lorentz from P and t.
I'll write P = (x,y,z), P' = (x',y',z'), K=(p,q,r),
w=(a,b,d), w'=(a',b',d').

Since P = wt + K, I have

x = at + p, y = bt + q, z = dt + r where a^2 + b^2 + d^2 = c^2.

In this form, the Lorentz is expressable as

x' = (x - vt) * G
y' = y
z' = z
t' = (t - vx/c^2) * G, where G = 1/sqrt(1-v^2/c^2)

which you've all seen before.

At this point, I need to reverse the transform; this is
easily enough done (I've done this myself in a post
some time back) and results in:

x = (x' + vt') * G
t = (t' + vx'/c^2) * G

The expressions will get very ugly otherwise, so I'm going to
change units early, forcing c = 1 and a^2+b^2+d^2 = 1. (This
is perfectly valid by using light-seconds instead of
meters, for example.)

The inverse Lorentz mutates into the rather pretty form

x = (x' + vt') * G
t = (t' + vx') * G

where G = 1/sqrt(1-v^2).

Since x = a*t+p,
(x' + v*t')*G = a*(t' + v*x')*G + p
(x' + v*t') = a*(t' + v*x') + p/G
(x' - a*v*x') = a*t' - v*t' + p/G
x'*(1 - a*v) = t'*(a-v) + p/G
x' = t'*(a-v)/(1-a*v) + p/(G * (1-a*v))

(Note the expression for a' here; it should look familiar. :-)
Since a is the component of the beam's velocity in the x direction,
this expression makes perfect sense.)

y=b*t+q implies

y' = b*G*(t'+v*x') + q
y' = b*G*(t'+t'*v*((a-v)/(1-a*v) + p/(G * (1-a*v))))+q
   = t'*b*G*(1 + v*((a-v)/(1-a*v))) + q'
   = t'*b*G*(1-a*v + a*v-v^2)/(1-a*v) + q'
   = t'*b*G*(1-v^2)/(1-a*v) + q'
   = t'*b/(G*(1-a*v)) + q'

   since G*(1-v^2) = 1/G.

I won't work out q' precisely as I'm more interested in
the velocity than the starting position of the ray; I
just know it contains no t' terms. Similarly,

z' = t'*d/(G*(1-a*v)) + r'

so a' = (a-v)/(1-a*v) = G*(a-v)/(G*(1-a*v)) ,
b' = b/(G*(1-a*v)) ,
d' = d/(G*(1-a*v)) .

making everything have a common denominator for simplicity.

Now (c')^2 = ||w'||^2 = (a')^2 + (b')^2 + (d')^2
= (G^2*(a^2+v^2-2*a*v)+b^2+d^2)/(G^2*(1 - a*v)^2)
= (G^2*(a^2+v^2-2*a*v)+1-a^2)/(G^2*(1 - a*v)^2)

(recall that a^2+b^2+d^2=1)

At this point, I divide by G^2/G^2, getting

(c')^2 = (a^2 + v^2 - 2*a*v + 1/G^2 - a^2/G^2)/(1 - a*v)^2

and substitute 1/G^2 = (1-v^2), and then grind:

(c')^2 = (a^2 + v^2 - 2*a*v + 1 - v^2 - a^2 * (1 - v^2))/(1 - a*v)^2
= (a^2 + v^2 - 2*a*v + 1 - v^2 - a^2 + a^2 * v^2)/(1 - a*v)^2
= (- 2*a*v + 1 + a^2 * v^2)/(1 - a*v)^2
= (1 - 2*a*v + a^2*v^2) / (1 - a*v)^2
= (1 - a*v)^2 / (1 - a*v)^2
= 1.

Therefore, this is a mathematical proof that c' = c for a light ray
going in an arbitrary direction, given a theoretical method
to measure OWLS, in flat space.

As it all should.

Of course math means little without experiment to back it up,
but one has to start somewhere. This also shows that MMX can
be moving through the luminiferous aether in any direction at
all, and still yield a null result.

>
> therefore SR is completely wrong from top to bottom, start to finish..

No, sorry, you're wrong. You're welcome to point out errors in
the math if you like, but I for one don't see any.

[rest snipped] 

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.

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