Re: Why TWLS=OWLS=c in any ONE Frame.

Henri Wilson wrote: 

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
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. 

Where did the 'absolute' come from?
What does it mean?

Paul
.

Relevant Pages

  • Re: Why TWLS=OWLS=c in any ONE Frame.
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  • Re: Why TWLS=OWLS=c in any ONE Frame.
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