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From: Tony M on 4 Aug 2010 21:25
On Aug 4, 9:13 pm, "Inertial" <relativ...(a)rest.com> wrote:
> "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

Here it is:

On Aug 4, 8:33 pm, Tony M <marc...(a)gmail.com> wrote:
> 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: rotchm on 4 Aug 2010 21:26

> > > 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?

Answer my above question. This question is alo the question that
Inertial is posing to you which you still have not answered. Rephrase
your scenario better.


> 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.

Correct.

>k
> is calculated as the ratio of the two frequencies, compensating for
> the classical Doppler shift, which leaves only the relativistic shift
> factor.

ok, so?
I define that Z is calculated as f' plus 5... What is your point?

From: Inertial on 4 Aug 2010 23:07
"Tony M" wrote in message
news:612f491e-7a64-438f-bfc9-23982699cac1(a)l14g2000yql.googlegroups.com...

On Aug 4, 9:13 pm, "Inertial" <relativ...(a)rest.com> wrote:
> "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

Here it is:

On Aug 4, 8:33 pm, Tony M <marc...(a)gmail.com> wrote:
> 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.

OK .. let see what you DID mean ...

> 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.

It depends on where and when the events are .. so k is not some constant for
all events for those two frames.

> Vice-versa, if O'
> measures an interval delta_t' then O must have measured
> delta_t=k*delta_t'.

If you're talking about the same pair of events, which generates a given k,
then that is trivially correct.

> 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: Tony M on 4 Aug 2010 23:24
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).

----------------------
Based on Inertial's excellent observation I am re-defining delta_t and
delta_t' as follows:

Let delta_t and delta_t’ be the time intervals measured by O and
respectively O' between two distinct events.
----------------------

> 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). It is unknown to the observers at this point whether delta_t>= delta_t’ or vice-versa and no assumption can be made (until they
>
> determine k). By their convention it can be one or the other, but not
> both. 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.
>
> The same ratio k applies to length and mass transformation. (Also, in
> this scenario the value of k has a special significance.)

From: Tony M on 4 Aug 2010 23:31
On Aug 4, 11:24 pm, Tony M <marc...(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).
>
> ----------------------
> Based on Inertial's excellent observation I am re-defining delta_t and
> delta_t' as follows:
>
> Let delta_t and delta_t’ be the time intervals measured by O and
> respectively O' between two distinct events.
> ----------------------
>
> > 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). It is unknown to the observers at this point whether delta_t>= delta_t’ or vice-versa and no assumption can be made (until they
>
> > determine k). By their convention it can be one or the other, but not
> > both. 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.
>
> > The same ratio k applies to length and mass transformation. (Also, in
> > this scenario the value of k has a special significance.)
>
>

I guess that's still not very clear. What I mean by two distinct
events is the same pair of distinct events for both observers, NOT
like someone said before: event one - O passes O' and event two - O'
passes O. That would be just one event.
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