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From: rotchm on 4 Aug 2010 20:08
On Aug 4, 5:47 pm, Tony M <marc...(a)gmail.com> wrote:
> Whether right or wrong, it’s just an idea, so here it is:
>
> Let there be two observers O and O’, moving directly towards each
> other at relative velocity v (considered positive in this case,
> negative if moving away). Let delta_t and delta_t’ be the rates
> measured by O and O’ on identical clocks at rest in their respective
> frames.

A little confusing here. Do you mean that delta_t is the rate of O as
measured by O and
delta_t' is the rate of O' as measured by O', or do you mean something
else?


>By convention both observers define k = delta_t/delta_t’

Ok, its a convention (definition).

> Now, the trick is to determine the value of k.
>
> One way of measuring k is by sending an EM signal from O to O’ and
> applying the formula k=f’/f/(1+v/c), where f and f’ are the
> frequencies of the EM signal as measured by O and O’ and then
> communicated to each other.

What do you mean that that is one way of measuring k !? :
-Where does that formula come from?
-Are you sure that that is a conventionnal way to measure what you
want to measure? Or, are you inventing your own rules for
measurements, rules which differ from SR's rules...?
-again, as for your delta_t _t', it is not clear what *you* mean by
your f and f'. Are they the observed frequency of the other clock
(signal generator), or do they represent the frequency at which one
receives the signal (the image, the crest of the EM, etc) ?


From: Tony M on 4 Aug 2010 20:33
On Aug 4, 7:24 pm, "Inertial" <relativ...(a)rest.com> wrote:
> "Tony M"  wrote in message
>
> news:5bc80553-c9ad-4c44-b7c8-c290d151dbda(a)f6g2000yqa.googlegroups.com...
>
>
>
> >Whether right or wrong, it’s just an idea, so here it is:
>
> >Let there be two observers O and O’, moving directly towards each
> >other at relative velocity v (considered positive in this case,
> >negative if moving away). Let delta_t and delta_t’ be the rates
> >measured by O and O’ on identical clocks at rest in their respective
> >frames.
>
> Every clock is measured to ticks at the correct rate in its own frame.  So
> delta_t and delta_t' are the same value.
>
> [snip rest]


Yes, you're right. They would both measure the same rate in their own
frame. I didn't formulate that one properly.
What I meant was that, by their convention, if O measures a time
interval delta_t, say between two events, he knows that O' has
measured a time interval delta_t'=delta_t/k. Vice-versa, if O'
measures an interval delta_t' then O must have measured
delta_t=k*delta_t'. So there is still no cross-frame measuring, O
measures delta_t and O' measures delta_t', and they both use the same
formula and factor for the transformation.
From: Androcles on 4 Aug 2010 20:35

"rotchm" <rotchm(a)gmail.com> wrote in message
news:9bd31262-8e1a-4a9d-baf4-18ee4dcc5a50(a)i31g2000yqm.googlegroups.com...
Or, are you inventing your own rules for
measurements, rules which differ from SR's rules...?
====================================
Why not? You wouldn't know the difference anyway.

From: Tony M on 4 Aug 2010 20:41
On Aug 4, 6:10 pm, dlzc <dl...(a)cox.net> wrote:
> Dear Tony M:
>
> On Aug 4, 2:47 pm, Tony M <marc...(a)gmail.com> wrote:
>
> > Whether right or wrong, it’s just an idea, so here
> > it is:
>
> > Let there be two observers O and O’, moving directly
> > towards each other at relative velocity v (considered
> > positive in this case, negative if moving away). Let
> > delta_t and delta_t’ be the rates measured by O and
> > O’ on identical clocks at rest in their respective
> > frames. By convention both observers define k =
> > delta_t/delta_t’ and agree on its value (where possible
> > values for k are 1/gamma <= k <= gamma).
>
> Where is the compensation for "classical" Doppler shift?
>
> With each tick, the other clock is closer (or farther) when it makes
> the next tick.
>
> David A. Smith

This part only sets the transformation convention between the two
frames, by using the same formula and same factor. There are no cross-
frame measurements. The two observers are not receiving ticks from
each-others clock.
The Doppler compensation takes place when they both measure the
frequency of the same monochromatic EM wave, say a laser beam, and
calculate k, since in this scenario the laser is being emitted by one
of the observers.
From: Inertial on 4 Aug 2010 20:54
"Tony M" wrote in message
news:3a88333c-6fdf-40d8-973f-2be25367ec20(a)14g2000yqa.googlegroups.com...
>This part only sets the transformation convention between the two
>frames, by using the same formula and same factor. There are no cross-
>frame measurements. The two observers are not receiving ticks from
>each-others clock.
>The Doppler compensation takes place when they both measure the
>frequency of the same monochromatic EM wave, say a laser beam, and
>calculate k, since in this scenario the laser is being emitted by one
>of the observers.

So as stated ... they BOTH MUST measure their own clocks as ticking at the
rate their own clocks tick (i.e. k = 1). So far your whole post has been
pointless.

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