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From: Tony M on 4 Aug 2010 20:55
On Aug 4, 8:08 pm, rotchm <rot...(a)gmail.com> wrote:
> 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) ?

There are no cross-frame measurements in this case. O shoots a laser
at local frequency f, O' receives the laser at his local frequency f'.
f' will carry both classical Doppler shift and relativistic shift. k
is calculated as the ratio of the two frequencies, compensating for
the classical Doppler shift, which leaves only the relativistic shift
factor.
From: Tony M on 4 Aug 2010 20:59
On Aug 4, 8:54 pm, "Inertial" <relativ...(a)rest.com> wrote:
> "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.

The reply to your comment was two posts above, you missed it.
From: Inertial on 4 Aug 2010 21:13
"Tony M" wrote in message
news:ba211296-0d67-44b7-bf03-d414b76d6341(a)i31g2000yqm.googlegroups.com...

On Aug 4, 8:54 pm, "Inertial" <relativ...(a)rest.com> wrote:
> "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.

> The reply to your comment was two posts above, you missed it.

Nope .. I'm looking at the thread here ... there is no reply to my previous
post in this thread.

If you have a reply, make it again.

So far your post is pointless.

Each observer observers his own rest clock ticking at the correct rate

From: Inertial on 4 Aug 2010 21:18
"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. 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).

So .. what you are saying (as written) is very plainly:

delta_t is the rate measured by O of a clock at rest wrt O
delta_t;;is the rate measured by O' of a clock at rest wrt O'

delta_t will be the same as delta_t'

if this is not the case .. please reword your scenario so it says what you
mean

From: Inertial on 4 Aug 2010 21:20
"Inertial" wrote in message
news:4c5a1177$0$11111$c3e8da3(a)news.astraweb.com...
[fix typo]
delta_t' is the rate measured by O' of a clock at rest wrt O'

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