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Inescapable (symmetric) twins paradox

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  Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 13:07 -0700
    Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2015-08-10 22:10 +0200
      Re: Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 13:36 -0700
        Re: Inescapable (symmetric) twins paradox Maciej Woźniak <mlwozniak@wp.pl> - 2015-08-10 22:54 +0200
    Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-08-10 16:53 -0500
      Re: Inescapable (symmetric) twins paradox Jon Price <jonelwoodprice@gmail.com> - 2015-08-10 18:24 -0700
      Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2015-08-10 23:06 -0700
      Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 05:43 -0700
      Re: Inescapable (symmetric) twins paradox RichD <r_delaney2001@yahoo.com> - 2015-09-01 09:32 -0700
        Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-09-02 10:32 -0500
          Re: Inescapable (symmetric) twins paradox RichD <r_delaney2001@yahoo.com> - 2015-09-06 22:01 -0700
            Re: Inescapable (symmetric) twins paradox Tom Roberts <tjroberts137@sbcglobal.net> - 2015-09-07 09:52 -0500
    Re: Inescapable (symmetric) twins paradox Koobee Wublee <koobee.wublee@gmail.com> - 2015-08-10 21:34 -0700
      Re: Inescapable (symmetric) twins paradox rotchm <rotchm@gmail.com> - 2015-08-11 07:39 -0700
        Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 08:06 -0700
          Re: Inescapable (symmetric) twins paradox rotchm <rotchm@gmail.com> - 2015-08-11 08:34 -0700
            Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 08:47 -0700
              Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-11 11:02 -0500
                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-11 14:49 -0700
                  Re: Inescapable (symmetric) twins paradox Gary Harnagel <hitlong@yahoo.com> - 2015-08-11 15:11 -0700
                  Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-11 17:16 -0500
                    Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 07:11 -0700
                      Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 09:53 -0500
                        Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 09:09 -0700
                          Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 11:29 -0500
                            Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 13:09 -0700
                              Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 15:39 -0500
                                Re: Inescapable (symmetric) twins paradox kenseto <setoken@att.net> - 2015-08-12 14:47 -0700
                                  Re: Inescapable (symmetric) twins paradox Odd Bodkin <bodkinodd@gmail.com> - 2015-08-12 16:52 -0500

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#360250 — Inescapable (symmetric) twins paradox

Hi Julio, 
In the experiment the clocks and twins change between different reference frames. For example, between earths frame and the one of motion away from earth. Since the clocks and twins change reference frames, doesn't that mean special relativity does not apply?

What is wrong with that argument/reasoning?

Jon

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#360252

Użytkownik "Jon Price"  napisał w wiadomości grup 
dyskusyjnych:5ab14ac8-8fd5-4790-bb42-e9825e8e8c8c@googlegroups.com...

|Hi Julio,
|In the experiment the clocks and twins change between different reference 
frames. For example, between earths frame and the one of motion away from 
earth. Since the clocks and twins change reference |frames, doesn't that 
mean special relativity does not apply?
|What is wrong with that argument/reasoning?

That "changing a reference frame" is just an insane mumble
without any real content.

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#360260

Why does a change in reference frames have no meaning?
Do reference frames have meaning?

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#360265

Użytkownik "Jon Price"  napisał w wiadomości grup 
dyskusyjnych:c543a88c-248b-4edf-bbfd-eb129e972d96@googlegroups.com...

|Why does a change in reference frames have no meaning?

And - what do You think it is? Physically.
It's not an acceleration. Acceleration is continous.
What else?
It's a pure fantasy.

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#360270

On 8/10/15 8/10/15 - 3:07 PM, Jon Price wrote:
> Hi Julio, In the experiment the clocks and twins change between different
> reference frames. For example, between earths frame and the one of motion
> away from earth. Since the clocks and twins change reference frames, doesn't
> that mean special relativity does not apply?

No. Not at all. But it DOES mean that simplistic applications of "time dilation" 
do not apply. One must CALCULATE the elapsed proper time for each twin via the 
appropriate integral over each path. This is straightforward, and when one does 
that, one finds that for the symmetric case of this thread the two twins' clocks 
display equal elapsed proper times. For the case you mention, assuming the 
"motion" twin comes back to earth to complete the scenario, then the "motion 
twin" will experience less elapsed proper time than the one who remained on 
earth [#].

	[#] Gravity, rotation, and the orbital velocity of the earth
	are all negligible in any example of the twin paradox in which
	the traveling twin leaves and comes back at a speed approaching
	c, so they are neglected and the earthbound twin is treated as
	inertial.



You seem to have jumped into the middle of a cesspool in which a number of 
idiots are attempting to argue with a few knowledgeable people (who are greatly 
outnumbered around here).

You may not have enough experience in physics to distinguish idiots and poseurs 
from people who understand modern physics. The only way I know to do that is 
that idiots and poseurs never reference textbooks, while knowledgeable people 
often do. I recommend:

	Taylor and Wheeler, _Spacetime_Physics_.


Tom Roberts

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#360290

Tom Roberts: Thanks for explanation and tip...

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#360308

On Monday, August 10, 2015 at 2:53:04 PM UTC-7, tjrob137 wrote:

> One must CALCULATE the elapsed proper time for each twin via the
> appropriate integral over each path.

If Tom understands the very basics, Tom will realize each transform relates two observations upon the same mutually observed.  In the case of the Twin paradox, it is the self-observation of one twin (A) compared with how the other twin (B) would observe this twin (A).  Of course, this is not complete.  We must find how the self-observation of B is observed by A.  Only from these two equations, we can relate how the rates of time flow go between A and B (each self-observation).  Again, this concept should be the basics.  <shrug>

Of course, this complete, professional analysis of the Twin paradox would spell out doom and disaster for the religion of SR.  After some brainstorming, these buffoons managed to create a manmade law of physics saying in non-inertial frames of reference, the observed rate of time flow at B (non-inertial) by A (inertial) becomes the actual local rate of time flow at A itself.  The other equation relating how the self-observation of B neatly ignored by the self-styled physicists.  To this day, there is no professional derivation how both equations derived would resolve the Twin paradox.  That is because the self-styled physicists are not engineers but armchair bullshitters.  <shrug>

So, beware of con-man's vocabulary "proper" since it can only mean the local geometry at the observed, and observations are merely projections like Tom has been preaching all along.  <shrug>

> [Tom's confusion, clarified by Koobee Wublee, snipped]

> You seem to have jumped into the middle of a cesspool in which a number of 
> idiots are attempting to argue with a few knowledgeable people (who are greatly 
> outnumbered around here).

Tom is an oppressor of truth but claims to be a victim of bulling by outnumbered dissenters of relativity.  That claim is just so ludicrous since the buffoons who are oblivious of what scientific method is manage to dominate in educating and brainwashing the younger generation.  <shrug>

> You may not have enough experience in physics to distinguish idiots and poseurs 
> from people who understand modern physics. The only way I know to do that is 
> that idiots and poseurs never reference textbooks, while knowledgeable people 
> often do.

Seen the movie "the Island"?  <shrug>

https://en.wikipedia.org/wiki/The_Island_(2005_film)

How can you tell a clone (impossible) from the real thing?  Don't let the buffooons do the thinkings for you, but let yourself do the thinking by understanding the subject.  Tom is totally pathetic when whining about been outnumbered.  <shrug>

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#360313

On Monday, August 10, 2015 at 5:53:04 PM UTC-4, tjrob137 wrote:
> On 8/10/15 8/10/15 - 3:07 PM, Jon Price wrote:
> > Hi Julio, In the experiment the clocks and twins change between different
> > reference frames. For example, between earths frame and the one of motion
> > away from earth. Since the clocks and twins change reference frames, doesn't
> > that mean special relativity does not apply?
> 
> No. Not at all. But it DOES mean that simplistic applications of "time dilation" 
> do not apply. One must CALCULATE the elapsed proper time for each twin via the 
> appropriate integral over each path. This is straightforward, and when one does 
> that, one finds that for the symmetric case of this thread the two twins' clocks 
> display equal elapsed proper times. For the case you mention, assuming the 
> "motion" twin comes back to earth to complete the scenario, then the "motion 
> twin" will experience less elapsed proper time than the one who remained on 
> earth [#].

Better interpretation is as follows:.
The "motion twin's" clock second is worth gamma second on the earth clock. In other words, a clock second is not a universal interval of time. This means that you can't compare the "motion twin's" clock second with the earth clock second directly. The GPS uses the above interpretation to synch the GPS clock with the earth clock. A GPS second is redefined to have 4.4647 more periods of the Cs 133 radiation than the standard ground clock second. This is designed to make the redefined GPS second contains the same amount of absolute time as the ground clock second and thus making the GPS in synch with the ground clock in terms of absolute time.


> 
> 	[#] Gravity, rotation, and the orbital velocity of the earth
> 	are all negligible in any example of the twin paradox in which
> 	the traveling twin leaves and comes back at a speed approaching
> 	c, so they are neglected and the earthbound twin is treated as
> 	inertial.
> 
> 
> 
> You seem to have jumped into the middle of a cesspool in which a number of 
> idiots are attempting to argue with a few knowledgeable people (who are greatly 
> outnumbered around here).
> 
> You may not have enough experience in physics to distinguish idiots and poseurs 
> from people who understand modern physics. The only way I know to do that is 
> that idiots and poseurs never reference textbooks, while knowledgeable people 
> often do. I recommend:
> 
> 	Taylor and Wheeler, _Spacetime_Physics_.
> 
> 
> Tom Roberts

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#362672

On August 10, 2015 tjrob137 wrote:
> One must CALCULATE the elapsed proper time for each twin via the 
> appropriate integral over each path. This is straightforward, 
> and when one does that, one finds that for the symmetric case 
> of this thread the two twins' clocks display equal elapsed proper 
> times. For the case you mention, assuming the "motion" twin 
> comes back to earth to complete the scenario, then the "motion 
> twin" will experience less elapsed proper time than the one who 
> remained on earth.

A thought experiment: consider a closed universe, 
as per GR, smaller than c/H, where H is Hubble's 
constant.  There, everything recedes at speed less than c.

Now, if the traveler is moving at constant speed close 
to c, in a straight line away from earth, he will 
forever remain a single inertial frame.  Eventually, 
he 'circles the universe', and returns to earth.

What does his clock read?

--
Rich

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#362762

[This is unrelated to the subject; you should have started a new thread.]

On 9/1/15 9/1/15 - 11:32 AM, RichD wrote:
> A thought experiment: consider a closed universe,
> as per GR, smaller than c/H, where H is Hubble's
> constant.  There, everything recedes at speed less than c.

Hmmmm. Not all closed universes have a Hubble constant. So let's just assume a 
traveler moving sufficiently fast can circumnavigate the universe.


> Now, if the traveler is moving at constant speed close
> to c, in a straight line away from earth, he will
> forever remain a single inertial frame.

Not really, because in such a universe there ARE no inertial frames. Remember 
"inertial frame" assumes it is of infinite extent and covers the universe, but:
   a) a closed universe admits no infinite frames, because the universe
      itself is of finite extent
   b) such a closed universe cannot be flat, and thus cannot be covered by
      a set of inertial coordinates.


> Eventually,
> he 'circles the universe', and returns to earth.
> What does his clock read?

Your description is woefully inadequate to determine the answer.



So let me discuss a situation that is similar to what you are apparently 
thinking of, for which an answer is known:

Consider a flat universe with spatial topology SxR^2 -- that is a cylinder with 
3 dimensions. Spacetime has topology SxR^2xR, and just like the usual 2-d 
cylinder SxR, it admits a flat metric.

Consider two twins, one who remains at home, and one who travels around the 
universe inertially, returning home. In a space-time diagram omitting the two R 
spatial dimensions, this universe is a cylinder, and we can select coordinates 
with the time axis running along the axis of the cylinder and the spatial 
coordinate going around the cylinder. In these coordinates, the stay-at-home 
twin's trajectory is parallel to the time axis along the cylinder, while the 
traveling twin's trajectory wraps once around the cylinder ("diagonally", along 
a helix), leaving and then rejoining the stay-at-home twin's trajectory. For 
this case, the traveling twin's elapsed proper time between meetings is less 
than that of the stay-at-home twin.


Tom Roberts

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#363331

 On September 2, tjrob137 wrote:
>> A thought experiment: consider a closed universe,
>> as per GR, smaller than c/H, where H is Hubble's
>> constant.  There, everything recedes at speed less than c.
>
> Not all closed universes have a Hubble constant. So let's just
> assume a traveler moving sufficiently fast can circumnavigate
> the universe.
>
>> Now, if the traveler is moving at constant speed close
>> to c, in a straight line away from earth, he will
>> forever remain a single inertial frame.
>
> Not really, because in such a universe there ARE no inertial
> frames. Remember "inertial frame" assumes it is of infinite
> extent and covers the universe, but:
>  a) a closed universe admits no infinite frames, because
> the universe itself is of finite extent
>  b) such a closed universe cannot be flat, and thus cannot
> be covered by a set of inertial coordinates.
>
>> Eventually, he 'circles the universe', and returns to earth.
>> What does his clock read?
>
> Your description is woefully inadequate to determine the answer.
> So let me discuss a situation that is similar to what you are
> apparently thinking of, for which an answer is known:
> Consider a flat universe with spatial topology SxR^2 --
> that is a cylinder with 3 dimensions. Spacetime has topology
> SxR^2xR, and just like the usual 2-d
> cylinder SxR, it admits a flat metric.
>
> Consider two twins, one who remains at home, and one who travels
> around the universe inertially, returning home. In a space-time
> diagram omitting the two R spatial dimensions, this universe is
> a cylinder, and we can select coordinates with the time axis
> running along the axis of the cylinder and the spatial
> coordinate going around the cylinder. In these coordinates,
> the stay-at-home twin's trajectory is parallel to the time axis
> along the cylinder, while the traveling twin's trajectory wraps
> once around the cylinder ("diagonally", along a helix), leaving
> and then rejoining the stay-at-home twin's trajectory. For
> this case, the traveling twin's elapsed proper time between
> meetings is less than that of the stay-at-home twin.

Interesting.  But if the traveler never experiences any acceleration, he remains in a single inertial frame 
throughout, how do we explain the result?

Also, in the simpler example I offered, a similar question: 
if the traveler never experiences acceleration, he's in 
free fall the whole time, why isn't he an inertial frame?

--
Rich

 

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#363355

On 9/7/15 9/7/15   12:01 AM, RichD wrote:
>   On September 2, tjrob137 wrote:
>>> A thought experiment: consider a closed universe,
>>> as per GR, smaller than c/H, where H is Hubble's
>>> constant.  There, everything recedes at speed less than c.
>>
>> Not all closed universes have a Hubble constant. So let's just
>> assume a traveler moving sufficiently fast can circumnavigate
>> the universe.
>>
>>> Now, if the traveler is moving at constant speed close
>>> to c, in a straight line away from earth, he will
>>> forever remain a single inertial frame.
>>
>> Not really, because in such a universe there ARE no inertial
>> frames. Remember "inertial frame" assumes it is of infinite
>> extent and covers the universe, but:
>>   a) a closed universe admits no infinite frames, because
>> the universe itself is of finite extent
>>   b) such a closed universe cannot be flat, and thus cannot
>> be covered by a set of inertial coordinates.
>>
>>> Eventually, he 'circles the universe', and returns to earth.
>>> What does his clock read?
>>
>> Your description is woefully inadequate to determine the answer.
>> So let me discuss a situation that is similar to what you are
>> apparently thinking of, for which an answer is known:
>> Consider a flat universe with spatial topology SxR^2 --
>> that is a cylinder with 3 dimensions. Spacetime has topology
>> SxR^2xR, and just like the usual 2-d
>> cylinder SxR, it admits a flat metric.
>>
>> Consider two twins, one who remains at home, and one who travels
>> around the universe inertially, returning home. In a space-time
>> diagram omitting the two R spatial dimensions, this universe is
>> a cylinder, and we can select coordinates with the time axis
>> running along the axis of the cylinder and the spatial
>> coordinate going around the cylinder. In these coordinates,
>> the stay-at-home twin's trajectory is parallel to the time axis
>> along the cylinder, while the traveling twin's trajectory wraps
>> once around the cylinder ("diagonally", along a helix), leaving
>> and then rejoining the stay-at-home twin's trajectory. For
>> this case, the traveling twin's elapsed proper time between
>> meetings is less than that of the stay-at-home twin.
>
> Interesting.  But if the traveler never experiences any acceleration, he remains in a single inertial frame
> throughout, how do we explain the result?

It is a topological difference.


> Also, in the simpler example I offered, a similar question:
> if the traveler never experiences acceleration, he's in
> free fall the whole time, why isn't he an inertial frame?

Because in your example there ARE no inertial frames.


Tom Roberts

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#360301

On Monday, August 10, 2015 at 1:07:05 PM UTC-7, Jon Price wrote:

> In the experiment the clocks and twins change between different
> reference frames. For example, between earths frame and the one of
> motion away from earth. Since the clocks and twins change reference
> frames, doesn't that mean special relativity does not apply?

The Lorentz transform, namely special relativity (SR), applies in both inertial and non-inertial frames of reference.  The stuff about non-inertial frames of reference where mathemaGics occurs is a manmade law of physics appended to SR.  Without this manmade law which cannot be found with the mathematical domains of SR, the Twin paradox can never be resolved.  <shrug>

https://groups.google.com/d/msg/sci.physics.relativity/9WNEIriVp5M/OhGBkODrYaoJ

> What is wrong with that argument/reasoning?

SR is a con-man's argument.  <shrug>

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#360321

On Tuesday, August 11, 2015 at 12:35:02 AM UTC-4, Koobee Wublee wrote:
 
> The Lorentz transform, namely special relativity (SR), applies 
> in both inertial and non-inertial frames of reference.  

Not true. Here is a counterexample of your claim;:

On Earth (a non constant grav field), place a clock at the bottom of a tall building and another at the top. The building will be your "spatial grid", your positions. The clocs\ks are at rest wrt each other. E-Synch your clock(s). Wait...wait...wait...

They will no longer be synchronized. That is an experimental fact. 

What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t, which contradicts our above conclusion. 

So you definately CANT apply the LT's as is from within an non inertial frame. 
You can however apply the relativity principle (as GR does) from within non iframes. 

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#360327

On Tuesday, August 11, 2015 at 10:39:36 AM UTC-4, rotchm wrote:
> On Tuesday, August 11, 2015 at 12:35:02 AM UTC-4, Koobee Wublee wrote:
>  
> > The Lorentz transform, namely special relativity (SR), applies 
> > in both inertial and non-inertial frames of reference.  
> 
> Not true. Here is a counterexample of your claim;:
> 
> On Earth (a non constant grav field), place a clock at the bottom of a tall building and another at the top. The building will be your "spatial grid", your positions. The clocs\ks are at rest wrt each other. E-Synch your clock(s). Wait...wait...wait...
> 
> They will no longer be synchronized. That is an experimental fact. 
> 
> What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t, which contradicts our above conclusion.

But v is not equal to zero. It is calculated as follows:
v=(universal wavelength of the source)(measured frequency of 
the source - measure frequency at the detector)

 
> 
> So you definately CANT apply the LT's as is from within an non inertial frame. 
> You can however apply the relativity principle (as GR does) from within non iframes.

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#360334

On Tuesday, August 11, 2015 at 11:06:42 AM UTC-4, kenseto wrote:

> > What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t, 
> >which contradicts our above conclusion.
> 
> But v is not equal to zero. It is calculated as follows:
> v=(universal wavelength of the source)(measured frequency of 
> the source - measure frequency at the detector)

But unfortunately for you, thats not how v (speed) is defined in SR. You are talking about something else. Stay on topic. [but you comment emphasizes the fact that position and time (and thus speed) have been defined in iframes. When not in iframes, one must specify the conventions used. You just proposed one, and I was proposing another; SR's. ]

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#360338

On Tuesday, August 11, 2015 at 11:34:55 AM UTC-4, rotchm wrote:
> On Tuesday, August 11, 2015 at 11:06:42 AM UTC-4, kenseto wrote:
> 
> > > What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t, 
> > >which contradicts our above conclusion.
> > 
> > But v is not equal to zero. It is calculated as follows:
> > v=(universal wavelength of the source)(measured frequency of 
> > the source - measure frequency at the detector)
> 
> But unfortunately for you, thats not how v (speed) is defined in SR.

SR is wrong and incomplete. That's why it is not capable of handling gravity. My formula is valid in all environments including gravity.


 You are talking about something else. Stay on topic. [but you comment emphasizes the fact that position and time (and thus speed) have been defined in iframes. When not in iframes, one must specify the conventions used. You just proposed one, and I was proposing another; SR's. ]

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#360341

On 8/11/2015 10:47 AM, kenseto wrote:
> On Tuesday, August 11, 2015 at 11:34:55 AM UTC-4, rotchm wrote:
>> On Tuesday, August 11, 2015 at 11:06:42 AM UTC-4, kenseto wrote:
>>
>>>> What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t,
>>>> which contradicts our above conclusion.
>>>
>>> But v is not equal to zero. It is calculated as follows:
>>> v=(universal wavelength of the source)(measured frequency of
>>> the source - measure frequency at the detector)
>>
>> But unfortunately for you, thats not how v (speed) is defined in SR.
>
> SR is wrong and incomplete. That's why it is not capable of handling gravity.
> My formula is valid in all environments including gravity.
>

Ken, you'd have to describe how your definition of v would help you 
measure the speed of a fastball at a baseball game.


-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#360393

On Tuesday, August 11, 2015 at 12:02:53 PM UTC-4, Odd Bodkin wrote:
> On 8/11/2015 10:47 AM, kenseto wrote:
> > On Tuesday, August 11, 2015 at 11:34:55 AM UTC-4, rotchm wrote:
> >> On Tuesday, August 11, 2015 at 11:06:42 AM UTC-4, kenseto wrote:
> >>
> >>>> What do the LT's predict: t' = (t-vx/c2)g. Since v = 0, t'=t,
> >>>> which contradicts our above conclusion.
> >>>
> >>> But v is not equal to zero. It is calculated as follows:
> >>> v=(universal wavelength of the source)(measured frequency of
> >>> the source - measure frequency at the detector)
> >>
> >> But unfortunately for you, thats not how v (speed) is defined in SR.
> >
> > SR is wrong and incomplete. That's why it is not capable of handling gravity.
> > My formula is valid in all environments including gravity.
> >
> 
> Ken, you'd have to describe how your definition of v would help you 
> measure the speed of a fastball at a baseball game.

You use the same formula I provided:
v=(universal wavelength of the radar souce)((initial frequency of the radar source - returning frequency)/2)

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#360394

On Tuesday, August 11, 2015 at 3:49:47 PM UTC-6, kenseto wrote:
>
> On Tuesday, August 11, 2015 at 12:02:53 PM UTC-4, Odd Bodkin wrote:
> >
> > Ken, you'd have to describe how your definition of v would help you 
> > measure the speed of a fastball at a baseball game.
> 
> You use the same formula I provided:
> v=(universal wavelength of the radar souce)((initial frequency of the
> radar source - returning frequency)/2)

There is no such thing as "universal wavelength."  Anyone with a brain
knows this, which explains why all you do is yammer, yammer, yammer.

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