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From: Tony M on 4 Aug 2010 17:47
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). 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: Androcles on 4 Aug 2010 18:07

"Tony M" <marcuac(a)gmail.com> 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). 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.

====================================================
The event of O' meeting O occurs before the event of O meeting O' because
less time has elapsed for O' than for O and "we cannot attach any absolute
signification to the concept of simultaneity, but that two events which,
viewed from a system of co-ordinates, are simultaneous, can no longer be
looked upon as simultaneous events when envisaged from a system which is in
motion relatively to that system. It is essential to have time defined by
means of stationary clocks in the stationary system, and the time now
defined being appropriate to the stationary system we call it ``the time of
the stationary system.'' -- Einstein.
Now, the trick is to determine which is the stationary lunatic, to hell with
the value of k.


From: dlzc on 4 Aug 2010 18:10
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
From: Inertial on 4 Aug 2010 19:24
"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]

From: harald on 4 Aug 2010 19:28
On Aug 4, 11:47 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). 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'll copy to you what I wasted on Seto; perhaps it is useful for you.
The Doppler equation ["classical" Doppler, thus not including k]:

fr = (c + vr)/(c - ve) * fe

[pay attention to the fact that it contains both vr AND ve.]

fr= f-received, fe = f-emitted,
c is propagation speed, vr= speed of receiver, ve = speed of emitter;
all speeds relative to the medium, with approaching speeds taken
positive.

Compare http://en.wikipedia.org/wiki/Doppler_effect

From that equation you can derive [slight rephrasing for you]:

1. what the pure Doppler effect is as measured in an assumed
"stationary" frame.
2. next you can calculate what the *apparent* Doppler effect is as
measured in a "moving frame", by observers who pretend to be
"stationary" while you assume that they are "moving".

You should find that the second case yields a false pure Doppler
effect:
it will appear for the moving observers as if the stationary clock
ticks slow by a factor gamma^2.

If you reached that, you will next be able to calculate in a few
seconds:

3. what the stationary observers will measure if the moving clock is
slow by a factor gamma, and
4. what in that case the moving observers will measure.

A few years ago I amused myself with such exercises, and found it very
helpful.

Success!
Harald
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