Re: Why TWLS=OWLS=c in any ONE Frame.
- From: "kenseto" <kenseto@xxxxxxxxxx>
- Date: Sun, 29 Jan 2006 15:46:31 GMT
"Henri Wilson" <HW@..> wrote in message
news:3v3ot1ple8nn8mdb1mg948lju9t3tjr340@xxxxxxxxxx
>
>
> A____________L______________B
>
> A and B are two observers equipped with light sources and clocks, at each
end
> of a long rigid rod of length L. Because the clocks are at mutually at
rest,
> the clock rates can be absolutely synched.
>
> When A's clock reads 0.000000000000, A sends a light pulse towards B. Its
speed
> is 'c' relative to A, and therefore also 'c' relative to B. Its travel
time is
> L/c (but at this stage we don't know the value of c).
>
> When the pulse arrives, B instantly adjusts his own clock to
0.0000000000000
This is not synchronizing the two clocks. A will read a greater time than B
at all time. If A abd B were correctly synchronized they will have the same
reading when compared to each other at all times. One way to correctly
synchronize Atwo clocks are as follows:
1. A and B are located in a common location.
2. Slow clock transport A and B in the opposite directions at the same
velocity and stop at the pre-measured locations (the same distance from the
common location).
3. A and B will remain synchronized according to all theories.
Using these two synchronized clock to measure OWLS will not give the value
c.
Ken Seto
> and immediately sends a return pulse towards A. It travels at 'c' wrt B
and
> therefore also 'c' wrt A. Its travel time is also L/c (but at this stage
we
> still don't know the value of c).
>
> This is sufficient to demonstrate that tAB=tBA. They are both numerically
equal
> to L/c. Only an aethrist would disagree.
>
> Next, A reads his clock when B's pulse arrives. It reads
> x.yxyeycjfmflfldfdgdkg.
>
> This reading, 't', is light's total travel time from A to B and back
again to
> A. From this A calculates c as being 2L/t.
>
> Note: A can replace B with a mirror and do the same thing, knowing that
the
> travel times in each direction must be the same.
>
> To synch the clocks A and B according to Einstein, B simply adjusts his
current
> clock reading upwards by (x.yxyeycjfmflfldfdgdkg)/2.
>
> Now, if A sends a pulse to B at time To, B receives it at (To+L/c), at
which
> instant A's clock must also read exactly (To+L/c).
>
> If B sends a pulse towards A at time T1, A receives it at (T1+L/c), at
which
> instant B's clock must also read exactly (T1+L/c)
>
> The clocks are now 'absolutely' synched.
>
> Poor old Einy didn't know what he actually achieved!!!!
>
>
> HW.
> www.users.bigpond.com/hewn/index.htm
>
>
.
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