Re: McCullough is caught with his pants down.
From: Androcles (Androcles_at_)
Date: 02/08/05
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Date: Tue, 08 Feb 2005 23:14:41 GMT
"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote in message
news:cubbjh011d@drn.newsguy.com...
> Androcles says...
>
>>"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote
>
>>> Androcles says...
>>>
>>>>Einstein defines it as (16+4)/2 = 16
>>>
>>> No, he does not.
>>
>>Yes, I've got it now.
>
> No, you don't.
Yes, I don't.
No, you do.
No, I do.
Yes, you don't.
>
>>He didn't say
>>"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. " when he said "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. "
>
> Where in the above did he say that he was defining time so that
> (16+4)/2 = 16? He didn't.
Well, see, the time for the mosquito (or light) to travel 80 units from
A (Sam) to
B (Joe) is 16 seconds, and the time for the mosquito to travel 20 units
from
B (Joe) to Sam (A) is 4 seconds, so we establish by definition that the
"time"
16 seconds equals the "time" 4 seconds.
Then he puts that in an equation, (tau0 + tau2) = tau1, or (16 +4)/4 =
16.
So I asked you why he did that, but now that I don't get it, you say he
didn't, which means he did. Or didn't, as the case may be.
Have you made that appointment with the psychiatrist yet?
> What he said was that as measured in the *moving* frame, the time
> required to travel from A to B is equal to the time required to
> travel from B to A. He didn't say that these times were equal
> as measured in the *stationary* frame.
Ah.... I see. Now it just so happens that some of your fellow psychotics
disagree with you on that, Because he didn't say "But the ray moves
relatively to the initial point of k, when measured in the stationary
system, with the velocity c-v, so that x'/(c-v) = t." and
c-v doesn't appear anywhere in his equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
and neither does ½, so he didn't say it (or did) and yes, I don't get it
(or no, I do).
>
> As I said before, if t_1 is the time at which the light signal
> leaves Sam, t_2 is the time at which it reaches Joe, and t_3
> the time in which the reflected signal reaches Sam, then we
> have for our example
>
> t_1 = 0
> t_2 = 16
> t_3 = 20
Yes, ducky.
(t_1 + t_3)/2 = t_2
only on Mondays, Wednesdays and Fridays, right?
>
> If t_1', t_2', and t_3' are the times of the same events, as
> measured in the moving frame, then for our example we have
>
> t_1' = 0
> t_2' = 8
> t_3' = 16
>
> Einstein's equation is about times in the moving frame.
Well, yeah, and the speed of light in the moving frame is c-v, c+v,
but he never said that. He said the speed of light was c in all inertial
frames, except he did because he never said, not once, anything like
t = x'/(c-v) or put that in an equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
that has a ½ at the front.
So you are right, I'm wrong, and no, I get it now (or yes, I don't, as
the case may be).
He's saying
> that
>
> t_2' = 1/2 (t_1' + t_3')
>
> which is true:
>
> 8 = 1/2 (0 + 16)
Yes, ducky, Joe sent the mosquito back to Origins instead of to Sam,
right?
Wanted to check on the hamburger flipping, I expect.
See a psychiatrist, McCullough, and while you are there make an
appointment
for Schwartz, moortel, Andersen, YBM, Draper, Poe et. al.
Get well soon, and come back and see me when you are no longer
confused about Newton's time being the same everywhere.
Good luck.
Androcles.
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