Why are the 'Fixed Stars' so FIXED?
"George Dishman" wrote in
:
"bz" wrote in message
98.139...
.....
You said:
E------------------D N (where N is slightly above or below the line of
sight from earth through D)
I say:
N+
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E------------------D
or |
N-
Or did I misunderstand what you said?
You are perfectly correct in your analysis but I think you
missed that I said the point of highest _acceleration_.
AH. I saw acceleration but wasn't sure WHAT was being accelerated. Light,
the Neutron star in its orbit, or ....?
I read it as 'rate of change of distance to earth' and didn't see how it
had anything to do with the [un]bunching of the pulses from the star,
which should be max at the points I said.
Two consecutive pulses emitted at points N+ and N- will
have almost identical velocities. I am saying the maximum
difference in velocities
[between consecutive pulses; maximum rate of change of pulse rate]
is when the positions are like this
~N+
E------------------D )
~ N-
The N+ pulse is emitted 1.5ms after the moment of alignment
and it then catches up to the earlier N- pulse which was
emitted 1.5ms before alignment.
The result is that the acceleration term is in quadrature
with the velocity term and the phase of the total relative
to the time of the peak of the Shapiro delay then places a
constraint on Henry's extinction distance.
Correct. The first derivative of the distance formula, from E to N will
give the rate of change of the distance and the derivative of sine is
cosine and vice versa so it will be 90 degrees out of phase.
Ok. We are in agreement.
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
remove ch100-5 to avoid spam trap
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