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correct transversal Doppler shift

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  correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-09 04:47 -0700
    Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-07-10 09:05 +0200
      Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-07-11 09:21 +0200
        Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-02 08:52 +0200
          Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-02 00:29 -0700
            Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-03 09:08 +0200
              Re: correct transversal Doppler shift Volney <volney@invalid.invalid> - 2022-08-03 04:33 -0400
                Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-04 08:01 +0200
                  Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 00:05 -0700
                    Re: correct transversal Doppler shift Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-04 00:11 -0700
                    Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-04 07:37 -0700
                      Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 08:04 -0700
                        Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-08-04 15:16 +0000
                        Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-04 08:21 -0700
                          Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-08-04 15:27 +0000
                          Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 08:59 -0700
                            Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-04 09:12 -0700
                    Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-05 07:08 +0200
                      Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-05 03:29 -0700
                        Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-06 10:35 +0200
                          Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-06 02:03 -0700
                            Re: correct transversal Doppler shift Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-06 02:26 -0700
                            Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-06 06:06 -0700
                              Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-07 07:18 +0200
                                Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-07 03:27 -0700
                                  Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-08 08:45 +0200
                                    Re: correct transversal Doppler shift "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2022-08-08 08:51 -0700
                                    Re: correct transversal Doppler shift Diego Traversa <toed@iavdattg.ev> - 2022-08-08 22:35 +0000
                                      Re: correct transversal Doppler shift "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2022-08-08 21:01 -0700
                      Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-05 03:33 -0700
                    Re: correct transversal Doppler shift "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2022-08-05 21:10 -0700
              Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-03 09:59 -0700
              Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-03 10:12 -0700
    Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-10 06:43 -0700
      Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-10 16:34 +0000
    Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-13 11:02 -0500
      Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-13 09:28 -0700
        Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-13 09:36 -0700
        Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-13 18:55 +0000
        Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-18 20:55 -0500
          Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-19 08:13 -0700
            Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-20 11:28 -0500
              Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-20 12:03 -0700
              Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-20 13:04 -0700
                Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-21 21:19 -0500
                  Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-22 07:28 -0700
                    Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-22 12:11 -0500
                    Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-22 17:19 +0000
            Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-22 22:44 +0000
              Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-23 04:46 -0700
                Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-23 04:59 -0700
                  Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-23 14:34 +0000
                Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-23 12:20 +0000
        Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-07-29 21:27 -0700
          Re: correct transversal Doppler shift Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-07-29 21:59 -0700
          Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-30 11:26 +0000
          Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-07-30 05:34 -0700
      Re: correct transversal Doppler shift Richard Hachel <r.hachel@tiscali.fr> - 2022-07-13 18:51 +0000
        Re: correct transversal Doppler shift Tom Roberts <tjroberts137@sbcglobal.net> - 2022-07-20 12:12 -0500
    Re: correct transversal Doppler shift "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2022-08-04 16:06 -0700
      Re: correct transversal Doppler shift Duan Bonomo <ddfs@dtszpmat.zy> - 2022-08-06 16:15 +0000
        Re: correct transversal Doppler shift Alsor <alsorgzl@gmail.com> - 2022-08-06 11:25 -0700
          Re: correct transversal Doppler shift Thomas Heger <ttt_heg@web.de> - 2022-08-07 07:25 +0200
    Re: correct transversal Doppler shift "erkd...@gmail.com" <erkdemon@gmail.com> - 2022-08-06 22:11 -0700

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#589350

On Saturday, August 6, 2022 at 3:35:49 AM UTC-5, Thomas Heger wrote:
> Am 05.08.2022 um 12:29 schrieb Prokaryotic Capase Homolog:
> > On Friday, August 5, 2022 at 12:08:33 AM UTC-5, Thomas Heger wrote: 
> >> Am 04.08.2022 um 09:05 schrieb Prokaryotic Capase Homolog: 
> >>> 
> >>> 1) Let us take a race car "r" traveling at Mach 0.2. 
> >>> 2) You, the observer "o", are standing one mile from the road. 
> >> No, I'm sitting twenty meters away from the track. 
> >>> 3) Sound travels 5 sec from the road to your ears at its closest point. 
> >>> 4) As the car crosses your path, you do not hear sound from where 
> >>> the car is NOW, but from where it was about 5 seconds ago "e". 
> >> I'm not a bat and cannot see by the ears. 
> >> Actually I see light and use my eyes for that purpose. 
> >
> > You do not hear with your ears. Stay on topic. 
> 
> I have always thought, that my ears are used for this purpose.

That was the "silly typo" that I mentioned in message
https://groups.google.com/g/sci.physics.relativity/c/z5sGxh8sv0s/m/cLq13x1CBQAJ
but canceling a message is never reliable.

> >> The delay for the finite speed of light needed to be considered, even if 
> >> very small. 
> >
> > For sonic transverse Doppler shift, it makes hardly any difference. 
> > 
> > For optical transverse Doppler shift, it is a different story. 
> 
> Why? 
> 
> Fast signals are also signals with finite speed. 
> 
> It is this finite speed, what causes the Doppler effect and not the 
> nature of the waves. 
> 
> So, why make a distinction between light and sound?

Sound is much slower and travels through a medium.
With sound, longitudinal Doppler effect is different for source moving
versus receiver moving.
With light, the same equation describes longitudinal Doppler effect for
source moving, receiver moving, or anything in between.

> >> But in this example it is totally irelevant, because the track does not 
> >> move and determins, where the race-car can go. 
> >>> 5) The sound that you hear from "e" is Doppler shifted to a higher tone. 
> >>> 
> >>> -------e--r---------- 
> >>> ----------|---------- 
> >>> ----------|---------- 
> >>> ----------|---------- 
> >>> ----------|---------- 
> >>> ----------|---------- 
> >>> ----------o---------- 
> >
> > In summary, whether the race car is *geometrically* at its 
> > closest point, or one *sees* the race car at its closest 
> > point, the overall effect is the same: One hears the sound 
> > of the race car Doppler-shifted to higher notes.
> > 
> >>> Let us try two more scenarios: 
> >>> 1) You are standing at the center of a circular track and a race car 
> >>> is driving around you at high speed. Is the sound that you hear 
> >>> Doppler shifted and if so, is it to a higher or lower tone? 
> >> I would prefer a linear track. 
> >> 
> >> The tone is higher on the side of the approach und lower on the receding 
> >> side, where the car drives away. 
> >
> > CIRCULAR track!!! At no point does the race car have any 
> > longitudinal motion with respect to you at the center. 
> > At race car speeds, one measures no significant sonic 
> > Doppler shift. 
> 
> Sure, circular track are a possibility. 
> 
> But the SRT setting does not contain anything similar to a circular race 
> track, but is restricted to streight lateral motion. 

Have you *NEVER* heard of Mossbauer rotor experiments, where 
source and detector are mounted in center and edge of rotor,
or vice versa?
https://en.wikipedia.org/wiki/Ives%E2%80%93Stilwell_experiment#M%C3%B6ssbauer_rotor_experiments

Besides which, it is absolutely *NOT TRUE* that SRT is applicable 
only to straight line motion. 
"An accelerated particle does not have an inertial frame in which it
is always at rest. However, an inertial frame can always be found 
which is momentarily comoving with the particle. This frame, the 
momentarily comoving reference frame (MCRF), enables application 
of special relativity to the analysis of accelerated particles."
https://en.wikipedia.org/wiki/Spacetime#Examples_of_4-vectors
Figure 4-3a is mine. Figure 4-3b is by somebody else.

> So, a simple street would be a better analog than a circular track.
> >>> 2) You are driving a race car around a circular track with a loudspeaker 
> >>> at the center. Is the sound that you hear Doppler shifted and if so, 
> >>> is it to a higher or lower tone? 
> >> No, I would always hear the same frequency, because the distance is the 
> >> important part and does not change in this example, if the speeds are 
> >> moderatly low. 
> >
> > Finally, you get one correct. At ordinary race car speeds, 
> > one measures no significant sonic Doppler shift in this 
> > scenario. 
> 
> 
> No! 
> 
> SONIC Doppler frequency shift is, of course, omnipresent in motor-sports. 

ARE YOU TOTALLY UNABLE TO READ AND UNDERSTAND THE
SCENARIO THAT I OUTLINED????

No *WONDER* you think that Einstein's 1905 paper is full of holes.
Your reading comprehension is abysmal.

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#589351

On Saturday, 6 August 2022 at 11:03:37 UTC+2, prokaryotic.c...@gmail.com wrote:

> Besides which, it is absolutely *NOT TRUE* that SRT is applicable 
> only to straight line motion. 

Surely it's not - even the hardest fanatics are not
stupid enough to really apply The Shit anywhere.

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#589362

sobota, 6 sierpnia 2022 o 11:03:37 UTC+2 prokaryotic.c...@gmail.com napisał(a):
> On Saturday, August 6, 2022 at 3:35:49 AM UTC-5, Thomas Heger wrote: 
> > Am 05.08.2022 um 12:29 schrieb Prokaryotic Capase Homolog: 
> > > On Friday, August 5, 2022 at 12:08:33 AM UTC-5, Thomas Heger wrote: 
> > >> Am 04.08.2022 um 09:05 schrieb Prokaryotic Capase Homolog: 
> > >>> 
> > >>> 1) Let us take a race car "r" traveling at Mach 0.2. 
> > >>> 2) You, the observer "o", are standing one mile from the road. 
> > >> No, I'm sitting twenty meters away from the track. 
> > >>> 3) Sound travels 5 sec from the road to your ears at its closest point. 
> > >>> 4) As the car crosses your path, you do not hear sound from where 
> > >>> the car is NOW, but from where it was about 5 seconds ago "e". 
> > >> I'm not a bat and cannot see by the ears. 
> > >> Actually I see light and use my eyes for that purpose. 
> > > 
> > > You do not hear with your ears. Stay on topic. 
> > 
> > I have always thought, that my ears are used for this purpose.
> That was the "silly typo" that I mentioned in message 
> https://groups.google.com/g/sci.physics.relativity/c/z5sGxh8sv0s/m/cLq13x1CBQAJ 
> but canceling a message is never reliable.
> > >> The delay for the finite speed of light needed to be considered, even if 
> > >> very small. 
> > > 
> > > For sonic transverse Doppler shift, it makes hardly any difference. 
> > > 
> > > For optical transverse Doppler shift, it is a different story. 
> > 
> > Why? 
> > 
> > Fast signals are also signals with finite speed. 
> > 
> > It is this finite speed, what causes the Doppler effect and not the 
> > nature of the waves. 
> > 
> > So, why make a distinction between light and sound?
> Sound is much slower and travels through a medium. 
> With sound, longitudinal Doppler effect is different for source moving 
> versus receiver moving. 
> With light, the same equation describes longitudinal Doppler effect for 
> source moving, receiver moving, or anything in between.

SR' equation describes something, but wrong.

See at the simple configuration:

....A
.../ v
./ 60 degs
O---B---->  v

Two bodies A and B start to move from point O, and move with the same speed, for example v =.6c,
and an angle between the both directions is 60 degs.

What is Doppler measured between A to B?
What is speed of A relative to B?
What speed provides radar measurement?

Where A is visible by B - what angle and speed?

We see: the speed A wrt B is just: 0.6c, because there is an equiangular triangle: 0.6 - 0.6 - 0.6

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#589439

Am 06.08.2022 um 15:06 schrieb Alsor:

>>>> For sonic transverse Doppler shift, it makes hardly any difference.
>>>>
>>>> For optical transverse Doppler shift, it is a different story.
>>>
>>> Why?
>>>
>>> Fast signals are also signals with finite speed.
>>>
>>> It is this finite speed, what causes the Doppler effect and not the
>>> nature of the waves.
>>>
>>> So, why make a distinction between light and sound?
>> Sound is much slower and travels through a medium.
>> With sound, longitudinal Doppler effect is different for source moving
>> versus receiver moving.
>> With light, the same equation describes longitudinal Doppler effect for
>> source moving, receiver moving, or anything in between.
>
> SR' equation describes something, but wrong.
>
> See at the simple configuration:
>
> ....A
> .../ v
> ./ 60 degs
> O---B---->  v
>
> Two bodies A and B start to move from point O, and move with the same speed, for example v =.6c,
> and an angle between the both directions is 60 degs.
>
> What is Doppler measured between A to B?
> What is speed of A relative to B?
> What speed provides radar measurement?

'Relativity' means 'relative to....'. Here the relation between A and B 
is important.

In your example A moves relative to B, because the points move along 
lines in an angle.

The 60° angle would make the geometric relations easy, because the 
points 0, A and B form a triangle with all angles having 60°.

Therefore the relative velocity between A and B should also be 0.6*c.


> Where A is visible by B - what angle and speed?

The angle between A and B is zero, because for A the only visible point 
is B.

The point 0 is not visible, hence the only possible reference is 
actually B and therefor the angle between B and line of sight is zero.

> We see: the speed A wrt B is just: 0.6c, because there is an equiangular triangle: 0.6 - 0.6 - 0.6


Sure..

But what was still your question?


TH

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#589468

niedziela, 7 sierpnia 2022 o 07:18:02 UTC+2 Thomas Heger napisał(a):
> Am 06.08.2022 um 15:06 schrieb Alsor: 
> 
> >>>> For sonic transverse Doppler shift, it makes hardly any difference. 
> >>>> 
> >>>> For optical transverse Doppler shift, it is a different story. 
> >>> 
> >>> Why? 
> >>> 
> >>> Fast signals are also signals with finite speed. 
> >>> 
> >>> It is this finite speed, what causes the Doppler effect and not the 
> >>> nature of the waves. 
> >>> 
> >>> So, why make a distinction between light and sound? 
> >> Sound is much slower and travels through a medium. 
> >> With sound, longitudinal Doppler effect is different for source moving 
> >> versus receiver moving. 
> >> With light, the same equation describes longitudinal Doppler effect for 
> >> source moving, receiver moving, or anything in between. 
> > 
> > SR' equation describes something, but wrong. 
> > 
> > See at the simple configuration: 
> > 
> > ....A 
> > .../ v 
> > ./ 60 degs 
> > O---B----> v 
> > 
> > Two bodies A and B start to move from point O, and move with the same speed, for example v =.6c, 
> > and an angle between the both directions is 60 degs. 
> > 
> > What is Doppler measured between A to B? 
> > What is speed of A relative to B? 
> > What speed provides radar measurement?
> 'Relativity' means 'relative to....'. Here the relation between A and B 
> is important. 
> 
> In your example A moves relative to B, because the points move along 
> lines in an angle. 
> 
> The 60° angle would make the geometric relations easy, because the 
> points 0, A and B form a triangle with all angles having 60°. 
> 
> Therefore the relative velocity between A and B should also be 0.6*c.
> > Where A is visible by B - what angle and speed?
> The angle between A and B is zero, because for A the only visible point 
> is B. 
> 
> The point 0 is not visible, hence the only possible reference is 
> actually B and therefor the angle between B and line of sight is zero.
> > We see: the speed A wrt B is just: 0.6c, because there is an equiangular triangle: 0.6 - 0.6 - 0.6
> Sure.. 
> 
> But what was still your question? 

My question is: 
- what speed, A wrt B, provides SR?
- what is measured Doppler here - from A to B?

And finally:
why SR provides v_AB <> 0.6c, 
and what is purpose of that velocity - where is the sense of that quantity?

Maybe: A moves relative to B with speed 0.6c,
but it doesn't move with speed 0.6c, but rather, say: 0.625c?

Is this the relativistic discovery - a great proposition?

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#589546

>>> See at the simple configuration:
>>>
>>> ....A
>>> .../ v
>>> ./ 60 degs
>>> O---B----> v
>>>
>>> Two bodies A and B start to move from point O, and move with the same speed, for example v =.6c,
>>> and an angle between the both directions is 60 degs.
>>>
>>> What is Doppler measured between A to B?
>>> What is speed of A relative to B?
>>> What speed provides radar measurement?
>> 'Relativity' means 'relative to....'. Here the relation between A and B
>> is important.
>>
>> In your example A moves relative to B, because the points move along
>> lines in an angle.
>>
>> The 60° angle would make the geometric relations easy, because the
>> points 0, A and B form a triangle with all angles having 60°.
>>
>> Therefore the relative velocity between A and B should also be 0.6*c.
>>> Where A is visible by B - what angle and speed?
>> The angle between A and B is zero, because for A the only visible point
>> is B.
>>
>> The point 0 is not visible, hence the only possible reference is
>> actually B and therefor the angle between B and line of sight is zero.
>>> We see: the speed A wrt B is just: 0.6c, because there is an equiangular triangle: 0.6 - 0.6 - 0.6
>> Sure..
>>
>> But what was still your question?
>
> My question is:
> - what speed, A wrt B, provides SR?
> - what is measured Doppler here - from A to B?
>
> And finally:
> why SR provides v_AB <> 0.6c,
> and what is purpose of that velocity - where is the sense of that quantity?

As I'm not particularily a fan of SRT, I cannot help you here.

You should ask one of the SRT-fans like e.g. Tom Roberts.


TH

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#589560

On Sunday, August 7, 2022 at 11:45:07 PM UTC-7, Thomas Heger wrote:
> >>> See at the simple configuration: 
> >>> 
> >>> ....A 
> >>> .../ v 
> >>> ./ 60 degs 
> >>> O---B----> v 
> >>> 
> >>> Two bodies A and B start to move from point O, and move with the same speed, for example v =.6c, 
> >>> and an angle between the both directions is 60 degs. 
> >>> 
> >>> What is Doppler measured between A to B? 
> >>> What is speed of A relative to B? 
> >>> What speed provides radar measurement? 
> >> 'Relativity' means 'relative to....'. Here the relation between A and B 
> >> is important. 
> >> 
> >> In your example A moves relative to B, because the points move along 
> >> lines in an angle. 
> >> 
> >> The 60° angle would make the geometric relations easy, because the 
> >> points 0, A and B form a triangle with all angles having 60°. 
> >> 
> >> Therefore the relative velocity between A and B should also be 0.6*c. 
> >>> Where A is visible by B - what angle and speed? 
> >> The angle between A and B is zero, because for A the only visible point 
> >> is B. 
> >> 
> >> The point 0 is not visible, hence the only possible reference is 
> >> actually B and therefor the angle between B and line of sight is zero. 
> >>> We see: the speed A wrt B is just: 0.6c, because there is an equiangular triangle: 0.6 - 0.6 - 0.6 
> >> Sure.. 
> >> 
> >> But what was still your question? 
> > 
> > My question is: 
> > - what speed, A wrt B, provides SR? 
> > - what is measured Doppler here - from A to B? 
> > 
> > And finally: 
> > why SR provides v_AB <> 0.6c, 
> > and what is purpose of that velocity - where is the sense of that quantity?
> As I'm not particularily a fan of SRT, I cannot help you here. 
> 
> You should ask one of the SRT-fans like e.g. Tom Roberts. 
> 
> 
> TH

I think it's more that Dr. Roberts defends and explicates SRT 
than necessarily, ascribing his opinion of it.

This is where he's familiar with both theories and their derivations, SR, STR.

This SRT I don't know what you're talking about, unless it's some camp of STR.

Both theories ascertain to the same objects, or their milieu.

Here's it's still "GR, then, SR, SR is local so massless light flees".

Instead of "STR, not necessarily GTR".

Basically this is for special relativity fundamental and special relativity applied, 
in the sense that it's for digital electronics the formalism to result shortest derivations.
(In STR, as far as junctions are concerned, while GTR is an overall larger concern.)

Of course, in geometric derivations for example there are all sorts derivations.

Just that "STR is less boilerplate and reduces few restrictions SR", 
still it's boilerplate including GTR again then gets much more, 
why and where whether SR and GR or STR and GTR are relevant.

Then this "GR, thus SR, ..., relativity fundamental", is where 
"relativity applied", STRA and GTRA, would be making for as 
simply "why and how applied".

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#589587

Thomas Heger wrote:

> As I'm not particularily a fan of SRT, I cannot help you here.
> You should ask one of the SRT-fans like e.g. Tom Roberts.

they did a big mistake in srt, thinking it was only about the speed. 
However, GR show its even worse being about the acceleration (gravity).

hence relativity is still incomplete, because beyond the acceleration we 
have jerks and rapidity.

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#589602

On Monday, August 8, 2022 at 3:35:04 PM UTC-7, Diego Traversa wrote:
> Thomas Heger wrote: 
> 
> > As I'm not particularily a fan of SRT, I cannot help you here. 
> > You should ask one of the SRT-fans like e.g. Tom Roberts.
> they did a big mistake in srt, thinking it was only about the speed. 
> However, GR show its even worse being about the acceleration (gravity). 
> 
> hence relativity is still incomplete, because beyond the acceleration we 
> have jerks and rapidity.

https://groups.google.com/g/sci.physics.relativity/search?q=acceleration%20author%3AFinlayson

"That's among reasons for a perspectiver [sic] about acceleration and
down from acceleration, about why velocity is accelerations' derivative
besides that it's changes' integral. "

"The "path integral" beyond "line integral" is sum-of-histories,
about this view from acceleration and its, "space derivative". "

This seems for a "sum of momentum histories".

Angular and linear, momentum, ....

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#589297

On Friday, August 5, 2022 at 12:08:33 AM UTC-5, Thomas Heger wrote:

I deleted a response with a silly typo, but I do not have time
to put together a corrected response. Some people on this
group are NOT retired. Wait until later.

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#589334

On Thursday, August 4, 2022 at 12:05:52 AM UTC-7, prokaryotic.c...@gmail.com wrote:
> On Thursday, August 4, 2022 at 1:01:24 AM UTC-5, Thomas Heger wrote: 
> 
> > The transversal Doppler effect for sound is therefor zero at the point, 
> > where the change of distance is momentarily zero. This point is the 
> > point, where the observer looks in right angles towards the racetrack.
> 1) Let us take a race car "r" traveling at Mach 0.2. 
> 2) You, the observer "o", are standing one mile from the road. 
> 3) Sound travels 5 sec from the road to your ears at its closest point. 
> 4) As the car crosses your path, you do not hear sound from where 
> the car is NOW, but from where it was about 5 seconds ago "e". 
> 5) The sound that you hear from "e" is Doppler shifted to a higher tone. 
> 
> -------e--r---------- 
> ----------|---------- 
> ----------|---------- 
> ----------|---------- 
> ----------|---------- 
> ----------|---------- 
> ----------o---------- 
> 
> Let us try two more scenarios: 
> 1) You are standing at the center of a circular track and a race car 
> is driving around you at high speed. Is the sound that you hear 
> Doppler shifted and if so, is it to a higher or lower tone? 
> 2) You are driving a race car around a circular track with a loudspeaker 
> at the center. Is the sound that you hear Doppler shifted and if so, 
> is it to a higher or lower tone?

Constant output:  here result with swing Doppler in sound, moving.

Doppler would start and end, but here the sound precedes the sound 
as even the sound builds and the rest of the Doppler in the slower sound, 
results according to the velocity and the much faster speed of sound.

Echo is about same, reflection that the mirror, is both Casimir, and, 
reflecting forward, that the sound travels.

High constant output / low constant output:  is acceleration in terms.

(And stop.)

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#589183

środa, 3 sierpnia 2022 o 09:08:06 UTC+2 Thomas Heger napisał(a):
> Am 02.08.2022 um 09:29 schrieb Prokaryotic Capase Homolog: 
> > On Tuesday, August 2, 2022 at 1:53:01 AM UTC-5, Thomas Heger wrote: 
> >> If you imagine the sound of a race car passing by on a track, you would 
> >> hear a certain distzinctive frequency change. 
> >> 
> >> It goes like 
> >> iiiieeoouuuuhh 
> >> 
> >> (something like that....) 
> >> 
> >> The 'blue-shift' side is, where the car approaches and the 'red-shift' 
> >> occurs, where the car recedes. 
> >> 
> >> But the 'transver case' is apparently coming with a zero frequency 
> >> shift, because that is the point, where frequency shift turns from red- 
> >> to blue-shift. 
> >> 
> >> The mid-point must be obviously zero-shift. 
> > 
> > Sloppy thinking, as usual. What is the "mid-point"?
> The meant 'mid-point' in a function graph is an extreme point, where the 
> tangent is horizontal. 
> 
> Now we take the change-of-frequency-function of the received sound of 
> the race car as function value and plot it on the y_axis. The distance 
> is plotted on the x-axis and the car paces from left to right. 
> 
> The mid point is now the point, where that path of the car is 
> perpendicular to my line of sight upon the race track. 
> 
> At this point the car moves perpendicular to my distance to the path, 
> what is called 'transversal movement'. 
> 
> Now we know from experience, that the Doppler effect at this point is 
> zero, because the 'change-of-frequency-function' is switching there 
> from positive to negative.
> > What is the red/blue Doppler shift of the race car when it is 
> > GEOMETRICALLY the nearest to you, the spectator?
> Zero 
> 
> It must be zero, because at this point the car changes from approaching 
> to receding. 
> 
> This is also a change from blue- to red-shift, or, if you prefer that, 
> from higher tone to lower tone.
> > What is the red/blue Doppler shift of the race car when you SEE it at 
> > its closest point?
> I would suggest to stick to sound. Light is also effected by the Doppler 
> effect, but it takes other means than ears to measure that.
> > Now apply this to a spaceship traveling at a large fraction of the speed 
> > of light.
> Spaceships cannot do this, because velocity requires a point of 
> reference, against which velocity is measured (assumed that space is 
> devoid of usable reference points). 
> 
> 
> TH

Light emitted by fast rocket has gamma time more energy.
so, the relativistic Doppler is still wrong,
because SR model assumes lower emitted of energy - gamma times.

k*k = k^2 = 1/1-v^2/c^2 -> this is an error of SR for Doppler measurements.

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#589184

On Wednesday, August 3, 2022 at 2:08:06 AM UTC-5, Thomas Heger wrote:
> Am 02.08.2022 um 09:29 schrieb Prokaryotic Capase Homolog: 

> > What is the red/blue Doppler shift of the race car when it is 
> > GEOMETRICALLY the nearest to you, the spectator?
> Zero 

I should have used an expression like "sonic Doppler shift", but you
seemed to understand my question correctly.

But your answer is wrong.

Read Volney's hints.
 
> It must be zero, because at this point the car changes from approaching 
> to receding. 

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#588197

sobota, 9 lipca 2022 o 13:47:57 UTC+2 Alsor napisał(a):
> The Doppler shift for angle 90 degs at receiver is blue not red: 
> 
> The general formula: 
> 
> d = k(1-v/c cosf); k = 1/sqrt(1-v^2/c^2) 
> 
> for transverse Doppler f = 90, cos90 = 0: 
> 
> d = k = 1/sqrt(1-v^2/c^2); 
> this is a bule shift! 
> 
> The energy of moving atom is equal: 
> E = m0 c^2 gamma 
> 
> Simple: the moving atoms emit bigger energy, 
> proportional to its total energy, which is gamma times bigger. 
> 
> 
> Where is the red shift? 
> 
> it's for angle cosf = v/c: 
> 
> d = k (1-v^2/c^2) = k/k^2 = 1/k = sqrt(1-v^2/c^2) 
> 
> ........... 
> 
> Total energy emitted by moving source is gamma times more, not less. 
> 
> forward + backward emission = 
> sqrt(1+v)/(1-v) + sqrt(1-v/1+v) = 
> sqrt(1-v^2) [1/1-v + 1/1+v] = 
> 
> sqrt(1-v^2) (1+v + 1-v)/(1-v^2) = 2/sqrt(1-v^2) 
> 
> the mean quantity is: 1/sqrt... 
> bigger gamma times, not less! 
> 
> There is no dilation of energy of moving sources - 
> the energy is bigger, always.

My intention is to notice:
the moving source has more energy, not less: 

E = mc^2 gamma; gamma > 1.

but the standard claims for many years the energy decrease,
E = mc^2 / gamma,

I just corrected this fallacy:

fast moving atoms emit more energy not less.

For transverse Doppler the frequency is bigger gamma times:
there is no time dilation of moving atom!
 
The 'time dilation' is for clocks - a cyclic process only, never for atomic emissions.

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#588210

Le 10/07/2022 à 15:43, Alsor a écrit :

 I have carefully read your comment which is very important.
 
> My intention is to notice:
> the moving source has more energy, not less: 

 You are right. 

> E = mc^2 gamma; gamma > 1.

 E=mc²/sqrt(1-v²/c²)=mc².sqrt(1+Vr²/c²)

 The moving source has more energy, not less. You are right. 

 
> 
> but the standard claims for many years the energy decrease,
> E = mc^2 / gamma,

 Warning! 

 gamma=1/sqrt(1-v²/c²)

> 
> I just corrected this fallacy:
> 
> fast moving atoms emit more energy not less.

 OK. 
> 
> For transverse Doppler the frequency is bigger

 It's true.

 
But when the orthogonal is drawn in the emitter frame.

We then have a deviation towards blue, more energetic.

The equation becomes:

L'=L.sqrt(1-v²/c²) if cosµ=0


But when the orthogonal is drawn in the emitter frame.

We then have a deviation towards blue, more energetic.

The equation becomes:

L'=L.sqrt(1-v²/c²) if cosµ=0 

<http://news2.nemoweb.net/jntp?fHOZ--k610AoLvkr9gXnC5Vy0PE@jntp/Data.Media:1>

But if the orthogonal us in R' (terrestrial frame) we have a déviation 
toward red. 

And another equation for R':

<http://news2.nemoweb.net/jntp?fHOZ--k610AoLvkr9gXnC5Vy0PE@jntp/Data.Media:2>

Less energetic. 

<http://news2.nemoweb.net/?DataID=fHOZ--k610AoLvkr9gXnC5Vy0PE@jntp>
R.H. 

-- 
"Mais ne nous trompons pas.
 Il n'y a pas que de la violence avec des armes : il y a des situations de 
violence." 
Abbé Pierre.

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#588370

On 7/9/22 6:47 AM, Alsor wrote:
> [...]

You were too sloppy in describing the physical situation you are
analyzing. Indeed you did not describe it at all -- you just stated some
equations without describing what their symbols actually mean.

Details matter.

The transverse Doppler shift is expected to be evaluated when the source
and receiver are at their point of closest approach, so the
instantaneous distance between them is not changing. For inertial source
and inertial receiver there are three possibilities:
  A) evaluate in the receiver frame. At the instant of closest
     approach, the light reaching the receiver was emitted
     earlier, when the source was approaching the receiver, so
     there is a blueshift.
  B) evaluate in the source frame. The light emitted at the
     point of closest approach reaches the receiver sometime
     later; there is a redshift.
  C) evaluate in the receiver frame, but consider the light
     emitted at the point of closest approach (that reaches
     the receiver sometime later); there is a redshift.

For both (B) and (C) the light ray being considered is seen by the
receiver as coming from the point of closest approach, redshifted. For
(A) the light ray being considered is seen by the receiver as coming
from a location before the point of closest approach, blueshifted.

As you did not describe the physical situation you are considering, it's
not obvious which of these you consider to be "transverse Doppler
shift", but your blueshift result is indicative.

Consider a laboratory measurement of transverse Doppler shift, with the
detector at rest in the lab and the source moving along some path;
the light path reaching the detector is shielded so only light
from the point of closest approach can reach the detector. Then (B)
and (C) apply [#] and SR predicts there to be a redshift.

    [#] The result is of course independent of which frame
    used for the analysis, as long as the physical situations
    are the same (same light ray is used).

For an astronomical measurement, it is quite difficult or impossible to
arrange similar constraints, because one rarely knows precisely where
the point of closest approach is located.

Tom Roberts

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#588371

środa, 13 lipca 2022 o 18:02:54 UTC+2 tjrob137 napisał(a):
> On 7/9/22 6:47 AM, Alsor wrote: 
> > [...] 
> 
> You were too sloppy in describing the physical situation you are 
> analyzing. Indeed you did not describe it at all -- you just stated some 
> equations without describing what their symbols actually mean. 
> 
> Details matter. 
> 
> The transverse Doppler shift is expected to be evaluated when the source 
> and receiver are at their point of closest approach, so the 
> instantaneous distance between them is not changing. For inertial source 
> and inertial receiver there are three possibilities: 
> A) evaluate in the receiver frame. At the instant of closest 
> approach, the light reaching the receiver was emitted 
> earlier, when the source was approaching the receiver, so 
> there is a blueshift. 
> B) evaluate in the source frame. The light emitted at the 
> point of closest approach reaches the receiver sometime 
> later; there is a redshift. 
> C) evaluate in the receiver frame, but consider the light 
> emitted at the point of closest approach (that reaches 
> the receiver sometime later); there is a redshift. 
> 
> For both (B) and (C) the light ray being considered is seen by the 
> receiver as coming from the point of closest approach, redshifted. For 
> (A) the light ray being considered is seen by the receiver as coming 
> from a location before the point of closest approach, blueshifted. 
> 
> As you did not describe the physical situation you are considering, it's 
> not obvious which of these you consider to be "transverse Doppler 
> shift", but your blueshift result is indicative. 
> 
> Consider a laboratory measurement of transverse Doppler shift, with the 
> detector at rest in the lab and the source moving along some path; 
> the light path reaching the detector is shielded so only light 
> from the point of closest approach can reach the detector. Then (B) 
> and (C) apply [#] and SR predicts there to be a redshift. 
> 
> [#] The result is of course independent of which frame 
> used for the analysis, as long as the physical situations 
> are the same (same light ray is used). 
> 
> For an astronomical measurement, it is quite difficult or impossible to 
> arrange similar constraints, because one rarely knows precisely where 
> the point of closest approach is located. 
> 
> Tom Roberts

I noticed only: the measured shift for moving and radiating atoms,
in the so called "transverse Doppler case", is blue not red,
because a moving atom has more energy: gamma times more, not less.

d = gamma x (1 - v/c cosf); 
this is for detector formula.

for f = 90, you get: gamma, not 1/gamma.

and this result is fully consistent with classics:
for 90 degs there is null Dopple, nothing changes!

Moving atom emits gamma more always (of energy or frequency: E = hf),
therefore for angle 90 degs we measure the same - what wos emitted: gamma hf.

And this is end of fallacy about time dilation of moving source.

The fantastic redshift is measured for:
f = cosf = -v/c,
so, now the source moves away at a little angle..

The source - an moving atom - emits at 90 degs,
but this is geometry... energy is still gamma times more,
as provided by the correct formula : E = gamma x mc^2.

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#588372

środa, 13 lipca 2022 o 18:28:59 UTC+2 Alsor napisał(a):
> środa, 13 lipca 2022 o 18:02:54 UTC+2 tjrob137 napisał(a): 
> > On 7/9/22 6:47 AM, Alsor wrote: 
> > > [...] 
> > 
> > You were too sloppy in describing the physical situation you are 
> > analyzing. Indeed you did not describe it at all -- you just stated some 
> > equations without describing what their symbols actually mean. 
> > 
> > Details matter. 
> > 
> > The transverse Doppler shift is expected to be evaluated when the source 
> > and receiver are at their point of closest approach, so the 
> > instantaneous distance between them is not changing. For inertial source 
> > and inertial receiver there are three possibilities: 
> > A) evaluate in the receiver frame. At the instant of closest 
> > approach, the light reaching the receiver was emitted 
> > earlier, when the source was approaching the receiver, so 
> > there is a blueshift. 
> > B) evaluate in the source frame. The light emitted at the 
> > point of closest approach reaches the receiver sometime 
> > later; there is a redshift. 
> > C) evaluate in the receiver frame, but consider the light 
> > emitted at the point of closest approach (that reaches 
> > the receiver sometime later); there is a redshift. 
> > 
> > For both (B) and (C) the light ray being considered is seen by the 
> > receiver as coming from the point of closest approach, redshifted. For 
> > (A) the light ray being considered is seen by the receiver as coming 
> > from a location before the point of closest approach, blueshifted. 
> > 
> > As you did not describe the physical situation you are considering, it's 
> > not obvious which of these you consider to be "transverse Doppler 
> > shift", but your blueshift result is indicative. 
> > 
> > Consider a laboratory measurement of transverse Doppler shift, with the 
> > detector at rest in the lab and the source moving along some path; 
> > the light path reaching the detector is shielded so only light 
> > from the point of closest approach can reach the detector. Then (B) 
> > and (C) apply [#] and SR predicts there to be a redshift. 
> > 
> > [#] The result is of course independent of which frame 
> > used for the analysis, as long as the physical situations 
> > are the same (same light ray is used). 
> > 
> > For an astronomical measurement, it is quite difficult or impossible to 
> > arrange similar constraints, because one rarely knows precisely where 
> > the point of closest approach is located. 
> > 
> > Tom Roberts
> I noticed only: the measured shift for moving and radiating atoms, 
> in the so called "transverse Doppler case", is blue not red, 
> because a moving atom has more energy: gamma times more, not less. 
> 
> d = gamma x (1 - v/c cosf); 
> this is for detector formula. 
> 
> for f = 90, you get: gamma, not 1/gamma. 
> 
> and this result is fully consistent with classics: 
> for 90 degs there is null Dopple, nothing changes! 
> 
> Moving atom emits gamma more always (of energy or frequency: E = hf), 
> therefore for angle 90 degs we measure the same - what wos emitted: gamma hf. 
> 
> And this is end of fallacy about time dilation of moving source. 
> 
> The fantastic redshift is measured for: 
> f = cosf = -v/c, 
> so, now the source moves away at a little angle.. 
> 
> The source - an moving atom - emits at 90 degs, 
> but this is geometry... energy is still gamma times more, 
> as provided by the correct formula : E = gamma x mc^2.

cosf = +v/c, not -v/c

then
doppler = gamma  ( 1 - v/c * v/c) = 1/gamma.

and this is not a transverse Doppler, and no any time dilation, but normal effect of recession;
because an angle is not eq. 90 degs, but more.

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#588382

Le 13/07/2022 à 18:28, Alsor a écrit :

> The source - an moving atom - emits at 90 degs,
> but this is geometry... energy is still gamma times more,
> as provided by the correct formula : E = gamma x mc^2.

Correct : blue shift.

I don't understand what's blocking Tom Roberts. 

R.H. 

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#588541

On 7/13/22 11:28 AM, Alsor wrote:
> I noticed only: the measured shift for moving and radiating atoms,
> in the so called "transverse Doppler case", is blue not red, because
> a moving atom has more energy: gamma times more, not less.

The energy of the source relative to the inertial frame of the detector
is not directly relevant. The source atoms emit photons with their usual
energy in the atoms' rest frame; to determine the energy seen by the
receiver requires one to use the relativistic Doppler shift equation,
not the formula for the energy of a moving object.

As I pointed out earlier in this thread, when one evaluates in the
receiver's inertial frame (to calculate what the receiver sees), and
considers photons emitted at the point of closest approach (which is
transverse in the receiver's inertial frame), there is a redshift.
Numerically:

    EnergySeenByReceiver / EnergyEmittedInEmitterFrame = 1/gamma

That is the same ratio as that for the "time dilation" of the moving
atom measured by the receiver's inertial frame.

Tom Roberts

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