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And Yet Another Reason Why the "medium" Matters Mr Andersen

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  And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-10-22 17:07 +0100
    Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-10-24 17:43 +0100
      Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-10-28 10:03 +0000
        Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-11-07 00:34 +0000
          Re: And Yet Another Reason Why the "medium" Matters Mr Andersen patdolan@comcast.net - 2018-11-07 03:55 -0800
            Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-11-07 12:05 +0000
              Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Ntowicz Usarczyk <iswso@wioioksi.io> - 2018-11-07 20:08 +0000
                Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-11-07 22:09 +0000
                  Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Gaston Castaneda <uucp@jav4ow.nq> - 2018-11-13 20:34 +0000
      Re: And Yet Another Reason Why the "medium" Matters Mr Andersen Keith Stein <keithstein111@gmail.com> - 2018-11-12 15:45 +0000

#475483 — And Yet Another Reason Why the "medium" Matters Mr Andersen

On 20/10/2018 09:21, Paul B. Andersen wrote:
 > Den 20.10.2018 04.52, skrev patdolan@comcast.net:
 >>
 >> a clock is traveling directly at me a .9c.  How much time do I see 
elapsed on the aforesaid clock for every hour that passes according to 
my wristwatch?
 >>
 >
 > If you are watching the approaching clock through a telescope,
 > you would for every hour on your clock see the approaching
 > clock advance:
 > According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
 > According to Newton: 1/(1-0.9) hours = 10 hours.


I rather think Newton never actually considered this problem,
Mr.Andersen, but assuming that your final line above means:

  According to CLASSICAL PHYSICS: 1/(1-0.9) hours = 10 hours.'

Then it's still wrong eh!

According to classical physics the problem CAN NOT BE SOLVED,me
UNLESS we know the frame of reference of the light medium eh!

Assuming the medium is stationary with respect to the Earth,
then your answer of 10 hours per hour is indeed correct for an
observer on Earth watching a clock in rocket approaching at .9c

BUT TD=10 is certainly NOT what the observer in the rocket approaching 
Earth at .9c would witness as he looks at Big Ben through his telescope.

So, still assuming the light medium is stationary with respect to the 
Earth, what TD would the observer in the rocket witness :
          a) According to SR ?
          b) According to Classical Physics ?

keith stein

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#475593

On 22/10/2018 17:07, Keith Stein wrote:
> 
> On 20/10/2018 09:21, Paul B. Andersen wrote:
>  > Den 20.10.2018 04.52, skrev patdolan@comcast.net:
>  >>
>  >> a clock is traveling directly at me a .9c.  How much time do I see 
> elapsed on the aforesaid clock for every hour that passes according to 
> my wristwatch?
>  >>
>  >
>  > If you are watching the approaching clock through a telescope,
>  > you would for every hour on your clock see the approaching
>  > clock advance:
>  > According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
>  > According to Newton: 1/(1-0.9) hours = 10 hours.
> 
> 
> I rather think Newton never actually considered this problem,
> Mr.Andersen, but assuming that your final line above means:
> 
>   According to CLASSICAL PHYSICS: 1/(1-0.9) hours = 10 hours.'
> 
> Then it's still wrong eh!
> 
> According to classical physics the problem CAN NOT BE SOLVED,me
> UNLESS we know the frame of reference of the light medium eh!
> 
> Assuming the medium is stationary with respect to the Earth,
> then your answer of 10 hours per hour is indeed correct for an
> observer on Earth watching a clock in rocket approaching at .9c
> 
> BUT TD=10 is certainly NOT what the observer in the rocket approaching 
> Earth at .9c would witness as he looks at Big Ben through his telescope.
> 
> So, still assuming the light medium is stationary with respect to the 
> Earth, what TD would the observer in the rocket witness :
>           a) According to SR ?
>           b) According to Classical Physics ?

I'd be waiting forever for an answer from you lot.
So, for the sake of completeness, i'll give the answer:

If you are watching "Big Ben" through a telescope as
you approach in your rocket at 0.9 c, then for every
hour on your clock you would see Big Ben advance

  According to SR: sqrt((1+0.9)/(1-0.9)) hours   = 4.36 hours
  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.
************************************************************
I'm confident that Mr.Andersen, who is actually pretty competent,
(for a relativist), could confirm that both those answers are
indeed correct, but he probably won't bother eh!

keith stein

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#475748

On 25/10/2018 17:16, Keith Stein wrote:
 > On 25/10/2018 14:59, Thomas 'PointedEars' Lahn wrote:
 >> Keith Stein wrote:

 >>> If you are watching "Big Ben" through a telescope,
 >>> as you approach in your rocket at 0.9, then for every
 >>> hour on your clock you would see Big Ben advance
 >>>  According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
 >>>  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.

 >> Ex falso quodlibet.  In this case, something ridiculously wrong. SR 
 >> says exactly the opposite: In the rocket frame, Big Ben’s clock
 >> would be observed to tick *slower* than the rocket clock (not
 >> faster).

  Thank you Thomas,
  Please write to Professor Andersen, TO LET HIM KNOW that, not only
  did he make a mistake with his  Classical Physics, but he doesn't
  understand Einstein's Special Relativity either. Alternatively
  you might be better advised to resit PHYSICS TEST II eh!  :)

keith stein

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#476460

On 28/10/2018 10:03, Keith Stein wrote:
> 
> On 25/10/2018 17:16, Keith Stein wrote:
>  > On 25/10/2018 14:59, Thomas 'PointedEars' Lahn wrote:
>  >> Keith Stein wrote:
> 
>  >>> If you are watching "Big Ben" through a telescope,
>  >>> as you approach in your rocket at 0.9, then for every
>  >>> hour on your clock you would see Big Ben advance
>  >>>  According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
>  >>>  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.
> 
>  >> Ex falso quodlibet.  In this case, something ridiculously wrong. SR 
>  >> says exactly the opposite: In the rocket frame, Big Ben’s clock
>  >> would be observed to tick *slower* than the rocket clock (not
>  >> faster).
> 
>   Thank you Thomas,
>   Please write to Professor Andersen, TO LET HIM KNOW that, not only
>   did he make a mistake with his  Classical Physics, but he doesn't
>   understand Einstein's Special Relativity either.


I am still waiting to see your letter to Professor Andersen, Thomas.
I need a good laugh eh!

keith stein

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#476487

On Tuesday, November 6, 2018 at 4:34:13 PM UTC-8, Keith Stein wrote:
> On 28/10/2018 10:03, Keith Stein wrote:
> > 
> > On 25/10/2018 17:16, Keith Stein wrote:
> >  > On 25/10/2018 14:59, Thomas 'PointedEars' Lahn wrote:
> >  >> Keith Stein wrote:
> > 
> >  >>> If you are watching "Big Ben" through a telescope,
> >  >>> as you approach in your rocket at 0.9, then for every
> >  >>> hour on your clock you would see Big Ben advance
> >  >>>  According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
> >  >>>  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.
> > 
> >  >> Ex falso quodlibet.  In this case, something ridiculously wrong. SR 
> >  >> says exactly the opposite: In the rocket frame, Big Ben’s clock
> >  >> would be observed to tick *slower* than the rocket clock (not
> >  >> faster).
> > 
> >   Thank you Thomas,
> >   Please write to Professor Andersen, TO LET HIM KNOW that, not only
> >   did he make a mistake with his  Classical Physics, but he doesn't
> >   understand Einstein's Special Relativity either.
> 
> 
> I am still waiting to see your letter to Professor Andersen, Thomas.
> I need a good laugh eh!
> 
> keith stein

Keith,

Professor Andersen is calculating the observed clock rate by means of the relativistic doppler equation whilst he is traveling *towards* Big Ben.  That is why he sees Ben ticking faster than his own clock EVEN THOUGH Ben is actually ticking 1/gamma slower than his clock.  If Andersen were traveling *away* from Ben at .9c then Ben would appear to him to be ticking even slower than a factor of 1/gamma, again because of the relativistic doppler shift.  This doppler slight of hand rescues the twin paradox. 

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#476488

On 07/11/2018 11:55, patdolan@comcast.net wrote:
> On Tuesday, November 6, 2018 at 4:34:13 PM UTC-8, Keith Stein wrote:
>> On 28/10/2018 10:03, Keith Stein wrote:
>>>
>>> On 25/10/2018 17:16, Keith Stein wrote:
>>>   > On 25/10/2018 14:59, Thomas 'PointedEars' Lahn wrote:
>>>   >> Keith Stein wrote:
>>>
>>>   >>> If you are watching "Big Ben" through a telescope,
>>>   >>> as you approach in your rocket at 0.9, then for every
>>>   >>> hour on your clock you would see Big Ben advance
>>>   >>>  According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
>>>   >>>  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.
>>>
>>>   >> Ex falso quodlibet.  In this case, something ridiculously wrong. SR
>>>   >> says exactly the opposite: In the rocket frame, Big Ben’s clock
>>>   >> would be observed to tick *slower* than the rocket clock (not
>>>   >> faster).
>>>
>>>    Thank you Thomas,
>>>    Please write to Professor Andersen, TO LET HIM KNOW that, not only
>>>    did he make a mistake with his  Classical Physics, but he doesn't
>>>    understand Einstein's Special Relativity either.
>>
>>
>> I am still waiting to see your letter to Professor Andersen, Thomas.
>> I need a good laugh eh!
>>
>> keith stein
> 
> Keith,
> 
> Professor Andersen is calculating the observed clock rate by means of the relativistic doppler equation whilst he is traveling *towards* Big Ben.  That is why he sees Ben ticking faster than his own clock EVEN THOUGH Ben is actually ticking 1/gamma slower than his clock.  If Andersen were traveling *away* from Ben at .9c then Ben would appear to him to be ticking even slower than a factor of 1/gamma, again because of the relativistic doppler shift.  This doppler slight of hand rescues the twin paradox.
> 

Thank you Pat. I do understand. It is PointedEars Lahn you should be 
explaining that to eh!

keith stein

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#476516

On Wed, 07 Nov 2018 12:05:39 +0000, Keith Stein wrote:

>> Professor Andersen is calculating the observed clock rate by means of
>> the relativistic doppler equation whilst he is traveling *towards* Big
>> Ben.  That is why he sees Ben ticking faster than his own clock EVEN
>> THOUGH Ben is actually ticking 1/gamma slower than his clock.  If
>> Andersen were traveling *away* from Ben at .9c then Ben would appear to
>> him to be ticking even slower than a factor of 1/gamma, again because
>> of the relativistic doppler shift.  This doppler slight of hand rescues
>> the twin paradox.
>
> Thank you Pat. I do understand. It is PointedEars Lahn you should be
> explaining that to eh!

What would be the difference between doppler shift and a relativistic 
doppler shift?

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#476522

On 07/11/2018 20:08, Ntowicz Usarczyk wrote:

> What would be the difference between doppler shift and a relativistic
> doppler shift?


-2.46 hours, Mr. Usarczyk.

 >> Keith Stein wrote:

  >>> If you are watching "Big Ben" through a telescope,
  >>> as you approach in your rocket at 0.9, then for every
  >>> hour on your clock you would see Big Ben advance
  >>>  According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
  >>>  According to Classical Physics: (1+.9)/1 hours = 1.9 hours.

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#476996

Keith Stein wrote:

> On 07/11/2018 20:08, Ntowicz Usarczyk wrote:
> 
>> What would be the difference between doppler shift and a relativistic
>> doppler shift?
> 
> -2.46 hours, Mr. Usarczyk.

Hours never writes that way, it says nothing. But hours, then minutes.

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#476870

On 24/10/2018 17:43, Keith Stein wrote:
> On 22/10/2018 17:07, Keith Stein wrote:
>>
>> On 20/10/2018 09:21, Paul B. Andersen wrote:
>>  > Den 20.10.2018 04.52, skrev patdolan@comcast.net:
>>  >>
>>  >> a clock is traveling directly at me a .9c.  How much time do I see 
>> elapsed on the aforesaid clock for every hour that passes according to 
>> my wristwatch?
>>  >>
>>  >
>>  > If you are watching the approaching clock through a telescope,
>>  > you would for every hour on your clock see the approaching
>>  > clock advance:
>>  > According to SR: sqrt((1+0.9)/(1-0.9)) hours = 4.36 hours
>>  > According to Newton: 1/(1-0.9) hours = 10 hours.
>>
>>
>> I rather think Newton never actually considered this problem,
>> Mr.Andersen, but assuming that your final line above means:
>>
>>   According to CLASSICAL PHYSICS: 1/(1-0.9) hours = 10 hours.'
>>
>> Then it's still wrong eh!
>>
>> According to classical physics the problem CAN NOT BE SOLVED,me
>> UNLESS we know the frame of reference of the light medium eh!
>>
>> Assuming the medium is stationary with respect to the Earth,
>> then your answer of 10 hours per hour is indeed correct for an
>> observer on Earth watching a clock in rocket approaching at .9c
>>
>> BUT TD=10 is certainly NOT what the observer in the rocket approaching 
>> Earth at .9c would witness as he looks at Big Ben through his telescope.
>>
>> So, still assuming the light medium is stationary with respect to the 
>> Earth, what TD would the observer in the rocket witness :
>>           a) According to SR ?
>>           b) According to Classical Physics ?
> 
> I'd be waiting forever for an answer from you lot.
> So, for the sake of completeness, i'll give the answer:
> 
> If you are watching "Big Ben" through a telescope as
> you approach in your rocket at 0.9 c, then for every
> hour on your clock you would see Big Ben advance
> 
>   According to SR: sqrt((1+0.9)/(1-0.9)) hours   = 4.36 hours
>   According to Classical Physics: (1+.9)/1 hours = 1.9 hours.
> ************************************************************
> I'm confident that Mr.Andersen, who is actually pretty competent,
> (for a relativist), could confirm that both those answers are
> indeed correct, but he probably won't bother eh!
> 
> keith stein
> 

Well at the very last point was SPOT ON.
But here is something which i think will elicit a response, from
my co-discovered of:

        THE         " ATKINSON-STEIN RELATIONSHIP "

      " SR prediction = The geometric mean of (Atkinson's) MAXIMUM
                                                         and
                                               (Stein's)   minimum"


I think we have stumbled upon some profound truth, of more general
application than the rocket approaching at ,9c which we were discussing
when this profound relationship of ours first emerged eh!

What do you think, Mr.(Prof?) Atkinson ? Shou;d we publish NOW, or
SHOULD WE AWAIT FURTHER CONFIRMATION ?
My own opinion is that we have 1 example. PUBLISH AND BE DAMNED  eh!
Is what i say, but you are the lead author, what do you think ?
  Mr.Atkinson ?

keith stein

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