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why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth?

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  why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-03-08 13:03 -0800
    Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Sam Wormley <swormley1@gmail.com> - 2016-03-08 16:03 -0600
      Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-09 22:00 -0500
        Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Sam Wormley <swormley1@gmail.com> - 2016-03-09 21:49 -0600
          Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-10 09:41 -0500
            Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-10 08:52 -0600
              Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-10 10:26 -0500
                Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-10 09:57 -0600
                  Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-10 16:42 -0500
                    Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-10 16:12 -0600
                      Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-10 22:21 -0500
                        Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-11 07:15 -0600
                          Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-11 12:18 -0500
                            Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-11 11:47 -0600
                              Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Jackpol11@hotmail.com - 2016-03-11 16:46 -0500
                                Re: why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Odd Bodkin <bodkinodd@gmail.com> - 2016-03-11 16:57 -0600
                            was not aware physicists use phony irrational numbers to compensate for LIGO Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-03-11 11:08 -0800
                              Re: was not aware physicists use phony irrational numbers to compensate for LIGO Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth? Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-03-12 16:10 -0800

#560494 — why did LIGO turn off their machine Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth?

On Monday, March 7, 2016 at 8:32:39 AM UTC-6, Odd Bodkin wrote:
Correct me if wrong, but this poster also seems to believe that the scientists of LIGO after their 15 or 14 Sept 2015 ping alleging 2 black holes, turned off their instruments and have been inactive since.

AP

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#560520

On 3/8/16 3:03 PM, Archimedes Plutonium wrote:
> On Monday, March 7, 2016 at 8:32:39 AM UTC-6, Odd Bodkin wrote:
> Correct me if wrong, but this poster also seems to believe that the scientists of LIGO after their 15 or 14 Sept 2015 ping alleging 2 black holes, turned off their instruments and have been inactive since.
>
> AP
>

   The detectors are running as we speak.

   Oh, and BTW, read this paper:
   Observation of Gravitational Waves from a Binary Black Hole Merger
 > 
http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102




-- 

sci.physics is an unmoderated newsgroup dedicated
to the discussion of physics, news from the physics
community, and physics-related social issues.

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#560999

On Tue, 8 Mar 2016 16:03:12 -0600, Sam Wormley <swormley1@gmail.com>
wrote:

>On 3/8/16 3:03 PM, Archimedes Plutonium wrote:
>> On Monday, March 7, 2016 at 8:32:39 AM UTC-6, Odd Bodkin wrote:
>> Correct me if wrong, but this poster also seems to believe that the scientists of LIGO after their 15 or 14 Sept 2015 ping alleging 2 black holes, turned off their instruments and have been inactive since.
>>
>> AP
>>
>
>   The detectors are running as we speak.
>
>   Oh, and BTW, read this paper:
>   Observation of Gravitational Waves from a Binary Black Hole Merger
> > 
>http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
That is a beautiful article but they had an equation there that makes
me think they are bluffing. The equation is for the chirp mass M. The
dimensions are all wrong in both versions.
To simplify, M to the 5th equals M1 M2 cubed divided by M1 plus M2 to
the 1/5 power. There's no way to make that into mass.
In a 2nd expression it begins with c^3/G which has units of kilograms
per second (a huge value). Therefore the following expression has to
boil down to time. But it can't, having frequency to 5/11 or whatever
it is, cannot possibly resolve into time.
This is fundamentally wrong and has to be explained.
John Polasek

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#561004

On 3/9/16 9:00 PM, Jackpol11@hotmail.com wrote:
> To simplify, M to the 5th equals M1 M2 cubed divided by M1 plus M2 to
> the 1/5 power. There's no way to make that into mass.


   Mass = (M1 M2)^3/5 / (M1 + M2)^1/5

   Works just fine, just algebra.


-- 

sci.physics is an unmoderated newsgroup dedicated
to the discussion of physics, news from the physics
community, and physics-related social issues.

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#561132

On Wed, 9 Mar 2016 21:49:37 -0600, Sam Wormley <swormley1@gmail.com>
wrote:

>On 3/9/16 9:00 PM, Jackpol11@hotmail.com wrote:
>> To simplify, M to the 5th equals M1 M2 cubed divided by M1 plus M2 to
>> the 1/5 power. There's no way to make that into mass.
>
>
>   Mass = (M1 M2)^3/5 / (M1 + M2)^1/5
>
>   Works just fine, just algebra.
The units don't match.
Mass = (M1 M2)/ (M1 + M2) is a proper expression = kg^2/kg = kg
The exponents must be the same for both terms but they have 3/5 and
1/5 resulting in
M = 5th root of kg^3/kg (=kg^2) = kg^2/5
Too much of "physics" is done with just algebra, but then it's not
physics. 
John Polasek

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#561137

On 3/10/2016 8:41 AM, Jackpol11@hotmail.com wrote:
> On Wed, 9 Mar 2016 21:49:37 -0600, Sam Wormley <swormley1@gmail.com>
> wrote:
>
>> On 3/9/16 9:00 PM, Jackpol11@hotmail.com wrote:
>>> To simplify, M to the 5th equals M1 M2 cubed divided by M1 plus M2 to
>>> the 1/5 power. There's no way to make that into mass.
>>
>>
>>    Mass = (M1 M2)^3/5 / (M1 + M2)^1/5
>>
>>    Works just fine, just algebra.
> The units don't match.
> Mass = (M1 M2)/ (M1 + M2) is a proper expression = kg^2/kg = kg
> The exponents must be the same for both terms but they have 3/5 and
> 1/5 resulting in
> M = 5th root of kg^3/kg (=kg^2) = kg^2/5
> Too much of "physics" is done with just algebra, but then it's not
> physics.
> John Polasek
>

John, you've made a mistake.
M1*M2 has dimensions [M]^2. This to the 3/5 power is [M] to the (6/5) power.
M1+M2 has dimensions [M]. This to the 1/5 power is [M] to the (1/5) power.
The quotient is [M] to the 5/5 power, or just [M].


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

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#561141

On Thu, 10 Mar 2016 08:52:30 -0600, Odd Bodkin <bodkinodd@gmail.com>
wrote:

>On 3/10/2016 8:41 AM, Jackpol11@hotmail.com wrote:
>> On Wed, 9 Mar 2016 21:49:37 -0600, Sam Wormley <swormley1@gmail.com>
>> wrote:
>>
>>> On 3/9/16 9:00 PM, Jackpol11@hotmail.com wrote:
>>>> To simplify, M to the 5th equals M1 M2 cubed divided by M1 plus M2 to
>>>> the 1/5 power. There's no way to make that into mass.
>>>
>>>
>>>    Mass = (M1 M2)^3/5 / (M1 + M2)^1/5
>>>
>>>    Works just fine, just algebra.
>> The units don't match.
>> Mass = (M1 M2)/ (M1 + M2) is a proper expression = kg^2/kg = kg
>> The exponents must be the same for both terms but they have 3/5 and
>> 1/5 resulting in
>> M = 5th root of kg^3/kg (=kg^2) = kg^2/5
>> Too much of "physics" is done with just algebra, but then it's not
>> physics.
>> John Polasek
>>
>
>John, you've made a mistake.
>M1*M2 has dimensions [M]^2. This to the 3/5 power is [M] to the (6/5) power.
>M1+M2 has dimensions [M]. This to the 1/5 power is [M] to the (1/5) power.
>The quotient is [M] to the 5/5 power, or just [M].
You're right, I forgot the square of the cube
M = 5th root of kg^3 NO,6 /kg (=kg^2).
I have trouble assimilating this, taking 5th roots of mass, though. 

Maybe you can explain their next expression that starts with c^3/G.
c^3/G = 4*10^35 kg/sec or 100,000 solar masses per second, which must
be multiplied by some value of time in the parenthesis again to give
M. What should be the value of that time and do the units match?
John Polasek

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#561150

On 3/10/2016 9:26 AM, Jackpol11@hotmail.com wrote:
> On Thu, 10 Mar 2016 08:52:30 -0600, Odd Bodkin <bodkinodd@gmail.com>

>
> Maybe you can explain their next expression that starts with c^3/G.
> c^3/G = 4*10^35 kg/sec or 100,000 solar masses per second, which must
> be multiplied by some value of time in the parenthesis again to give
> M. What should be the value of that time and do the units match?
> John Polasek
>

Well of course the units match. As you say, c^3/G has units [M]/[T], so 
if you multiply it by a time you get something with units [M].


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

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#561277

On Thu, 10 Mar 2016 09:57:05 -0600, Odd Bodkin <bodkinodd@gmail.com>
wrote:

>On 3/10/2016 9:26 AM, Jackpol11@hotmail.com wrote:
>> On Thu, 10 Mar 2016 08:52:30 -0600, Odd Bodkin <bodkinodd@gmail.com>
>
>>
>> Maybe you can explain their next expression that starts with c^3/G.
>> c^3/G = 4*10^35 kg/sec or 100,000 solar masses per second, which must
>> be multiplied by some value of time in the parenthesis again to give
>> M. What should be the value of that time and do the units match?
>> John Polasek
>>
>
>Well of course the units match. As you say, c^3/G has units [M]/[T], so 
>if you multiply it by a time you get something with units [M].
You accept it without challenging it.
I had my doubts about making the content into time but the bracket
simplified is 
     (f^-11/3*fdot)^3/5  comes out as seconds.
I just haven't had that much contact with frequency to the -11/3 or
the 5th root of mass cubed, to be comfortable with it. (Your intuition
may vary) 
c^3/G x bracket time:
The bracket time should be 620us with c^3/G at 100,000 Msol/sec with
final mass 62Msol from Table I:
  T = 62/10^5 or 620/10^6, or 620 usec
62 Msol comes and goes in less than a msec. Time to try to see what
the physics is here. Just wondering.
John Polasek

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#561287

On 3/10/2016 3:42 PM, Jackpol11@hotmail.com wrote:
> On Thu, 10 Mar 2016 09:57:05 -0600, Odd Bodkin <bodkinodd@gmail.com>

>> Well of course the units match. As you say, c^3/G has units [M]/[T], so
>> if you multiply it by a time you get something with units [M].
> You accept it without challenging it.
> I had my doubts about making the content into time but the bracket
> simplified is
>       (f^-11/3*fdot)^3/5  comes out as seconds.
> I just haven't had that much contact with frequency to the -11/3 or
> the 5th root of mass cubed, to be comfortable with it. (Your intuition
> may vary)
> c^3/G x bracket time:
> The bracket time should be 620us with c^3/G at 100,000 Msol/sec with
> final mass 62Msol from Table I:
>    T = 62/10^5 or 620/10^6, or 620 usec
> 62 Msol comes and goes in less than a msec. Time to try to see what
> the physics is here. Just wondering.
> John Polasek
>

I'll help you see where this comes from.
f has units 1/[T].
f-dot has units 1/[T]^2, because it's the time derivative of the frequency.
So the product inside the brackets has units [T] to the 11/3 power 
divided by [T]^2. So that power is 11/3 - 6/3 = 5/3.
Now you take that quantity to the 3/5 power, leaving you with [T] to the 
(5/3) x (3/5) power or just [T].

It's not difficult.


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

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#561385

On Thu, 10 Mar 2016 16:12:30 -0600, Odd Bodkin <bodkinodd@gmail.com>
wrote:

>On 3/10/2016 3:42 PM, Jackpol11@hotmail.com wrote:
>> On Thu, 10 Mar 2016 09:57:05 -0600, Odd Bodkin <bodkinodd@gmail.com>
>
>>> Well of course the units match. As you say, c^3/G has units [M]/[T], so
>>> if you multiply it by a time you get something with units [M].
>> You accept it without challenging it.
>> I had my doubts about making the content into time but the bracket
>> simplified is
>>       (f^-11/3*fdot)^3/5  comes out as seconds.
>> I just haven't had that much contact with frequency to the -11/3 or
>> the 5th root of mass cubed, to be comfortable with it. (Your intuition
>> may vary)
>> c^3/G x bracket time:
>> The bracket time should be 620us with c^3/G at 100,000 Msol/sec with
>> final mass 62Msol from Table I:
>>    T = 62/10^5 or 620/10^6, or 620 usec
>> 62 Msol comes and goes in less than a msec. Time to try to see what
>> the physics is here. Just wondering.
>> John Polasek
>>
>
>I'll help you see where this comes from.
>f has units 1/[T].
>f-dot has units 1/[T]^2, because it's the time derivative of the frequency.
>So the product inside the brackets has units [T] to the 11/3 power 
>divided by [T]^2. So that power is 11/3 - 6/3 = 5/3.
>Now you take that quantity to the 3/5 power, leaving you with [T] to the 
>(5/3) x (3/5) power or just [T].
>
>It's not difficult.
For reference see:
http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
I didn't say it was difficult-I said it was unspeakable. I worked out
the units myself as you saw in the previous note so that's behind us.
A bastard term like frequency to the -11/3 power needs someone to
explain its relevance and its origin. Or at least give it units so
that it should be understood. for
 Let L = wavelength or m/cycle = 1/F. Then we have
     f^-11/3 = L^11/3 = cube root of (meters^11/cycle^11)
This is hard to believe
Things don't really improve by taking the 3/5 power of all that. 
This algebra is really dubious.
In the other note, I was looking for some enlightenment about 62 solar
masses coming and going in less than a millisecond, as I figured out
in the last note.
John Polasek

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#561490

On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
> For reference see:
> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
> I didn't say it was difficult-I said it was unspeakable. I worked out
> the units myself as you saw in the previous note so that's behind us.
> A bastard term like frequency to the -11/3 power needs someone to
> explain its relevance and its origin.

The derivation of that equation can be found in the article referenced 
in this paper. If you would bother to look that up, you might have your 
questions answered.
L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
This is the reason the reference is listed, so that you not think it was 
made up from thin air.


> Or at least give it units so
> that it should be understood. for
>   Let L = wavelength or m/cycle = 1/F. Then we have
>       f^-11/3 = L^11/3 = cube root of (meters^11/cycle^11)

I really don't think that's necessary. You're asking scientists to 
explain the units in a way that is understandable TO YOU. But as you can 
see, even others on this group have not struggled with it to the degree 
you have, and I'm fairly confident physicists do not struggle with it.

> This is hard to believe
> Things don't really improve by taking the 3/5 power of all that.

I don't know what you mean by "improve". The theory is the theory.

> This algebra is really dubious.
> In the other note, I was looking for some enlightenment about 62 solar
> masses coming and going in less than a millisecond, as I figured out
> in the last note.
> John Polasek


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

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#561589

On Fri, 11 Mar 2016 07:15:19 -0600, Odd Bodkin <bodkinodd@gmail.com>
wrote:

>On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
>> For reference see:
>> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
>> I didn't say it was difficult-I said it was unspeakable. I worked out
>> the units myself as you saw in the previous note so that's behind us.
>> A bastard term like frequency to the -11/3 power needs someone to
>> explain its relevance and its origin.
>
>The derivation of that equation can be found in the article referenced 
>in this paper. If you would bother to look that up, you might have your 
>questions answered.
>L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
>Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
>This is the reason the reference is listed, so that you not think it was 
>made up from thin air.
It's unavailable without a subscription.

>
>> Or at least give it units so
>> that it should be understood. for
>>   Let L = wavelength or m/cycle = 1/F. Then we have
>>       f^-11/3 = L^11/3 = cube root of (meters^11/cycle^11)
>
>I really don't think that's necessary. You're asking scientists to 
>explain the units in a way that is understandable TO YOU. 
No I am not. The units are where the physics is. Just making equations
balance is not physics, if the equations are not the logical result of
quantizing some physical sequence. f to the -11/3 is an example
Of the 118 references in the paper, #11 was not available without a
subscription. 

>> Things don't really improve by taking the 3/5 power of all that.
>
>I don't know what you mean by "improve". The theory is the theory.

I don't see a theory, I see an assertion. I see a believer, or no, a
person who is unwilling to doubt. 
There is another paper that gives a lot more information to the inside
workings:
http://arxiv.org/pdf/1602.03840v1.pdf
It's hard to see how they identified the 2 masses as 30 and 32 solar
masses, when the 1st notification was on earth at the time of the
collision 1.3 Byrs later and had no idea in what direction to look
(for 2 black holes).
This is much like supernovae with a life several weeks, which appear
at random, the best of which are identified as "no more than 5 days
after the peak." Not one has been tracked from start to finish, so
there is no functional form to describe the "light curve". it is
necessary to "back-engineer" to the assumed peak. 
(But easier than tracking those 2 black holes after they are
demolished).

John Polasek

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#561593

On 3/11/2016 11:18 AM, Jackpol11@hotmail.com wrote:
> On Fri, 11 Mar 2016 07:15:19 -0600, Odd Bodkin <bodkinodd@gmail.com>
> wrote:
>
>> On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
>>> For reference see:
>>> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
>>> I didn't say it was difficult-I said it was unspeakable. I worked out
>>> the units myself as you saw in the previous note so that's behind us.
>>> A bastard term like frequency to the -11/3 power needs someone to
>>> explain its relevance and its origin.
>>
>> The derivation of that equation can be found in the article referenced
>> in this paper. If you would bother to look that up, you might have your
>> questions answered.
>> L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
>> Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
>> This is the reason the reference is listed, so that you not think it was
>> made up from thin air.
> It's unavailable without a subscription.

Whose problem is that supposed to be?
You have options:
1. You purchase the article.
2. You visit a university library that has purchased a subscription.

One involves money, one involves legwork. Which of these are you willing 
to invest?

>
>>
>>> Or at least give it units so
>>> that it should be understood. for
>>>    Let L = wavelength or m/cycle = 1/F. Then we have
>>>        f^-11/3 = L^11/3 = cube root of (meters^11/cycle^11)
>>
>> I really don't think that's necessary. You're asking scientists to
>> explain the units in a way that is understandable TO YOU.
> No I am not. The units are where the physics is. Just making equations
> balance is not physics, if the equations are not the logical result of
> quantizing some physical sequence. f to the -11/3 is an example
> Of the 118 references in the paper, #11 was not available without a
> subscription.

I don't think physicists owe laypeople an explanation in their papers 
how to understand the physics of those equations in terms of dimensions. 
That's where you're expected to do the background reading, and that's 
why they include references. Scientific papers are NOT SELF-CONTAINED, 
where everything you'd want to know about the subject is wholly 
contained in the paper.

>
>>> Things don't really improve by taking the 3/5 power of all that.
>>
>> I don't know what you mean by "improve". The theory is the theory.
>
> I don't see a theory, I see an assertion. I see a believer, or no, a
> person who is unwilling to doubt.

Sorry, but to believe, you really need to read the references too.

> There is another paper that gives a lot more information to the inside
> workings:
> http://arxiv.org/pdf/1602.03840v1.pdf
> It's hard to see how they identified the 2 masses as 30 and 32 solar
> masses, when the 1st notification was on earth at the time of the
> collision 1.3 Byrs later and had no idea in what direction to look
> (for 2 black holes).

How they derived the two masses is explained in the paper. It's solving 
for M1 and M2 from the equations given. They DO NOT NEED to show you all 
the steps.

> This is much like supernovae with a life several weeks, which appear
> at random, the best of which are identified as "no more than 5 days
> after the peak." Not one has been tracked from start to finish, so
> there is no functional form to describe the "light curve". it is
> necessary to "back-engineer" to the assumed peak.
> (But easier than tracking those 2 black holes after they are
> demolished).
>
> John Polasek
>


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

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#561694

On Fri, 11 Mar 2016 11:47:57 -0600, Odd Bodkin <bodkinodd@gmail.com>
wrote:

>On 3/11/2016 11:18 AM, Jackpol11@hotmail.com wrote:
>> On Fri, 11 Mar 2016 07:15:19 -0600, Odd Bodkin <bodkinodd@gmail.com>
>> wrote:
>>
>>> On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
>>>> For reference see:
>>>> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
>>>> I didn't say it was difficult-I said it was unspeakable. I worked out
>>>> the units myself as you saw in the previous note so that's behind us.
>>>> A bastard term like frequency to the -11/3 power needs someone to
>>>> explain its relevance and its origin.
>>>
>>> The derivation of that equation can be found in the article referenced
>>> in this paper. If you would bother to look that up, you might have your
>>> questions answered.
>>> L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
>>> Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
>>> This is the reason the reference is listed, so that you not think it was
>>> made up from thin air.
>> It's unavailable without a subscription.
>
>Whose problem is that supposed to be?
>You have options:
>1. You purchase the article.
>2. You visit a university library that has purchased a subscription.
>
>One involves money, one involves legwork. Which of these are you willing 
>to invest?
I think you're mistaking my volubility for fervor. I'm just looking
for someone to explain the term frequency to the -11/3 power,
augmented by 3/5  power, defining a unit of time, as the cofactor to 
c^3/G. But how do I know, out of 118 papers, that one of them will
actually come across? I am not willing to gamble $35. 
I evaluated the equation coming up with 620 µs for 32Msun. What does
that time really represent?
Let's hope they stay tuned for the next calamity. 
(Black holes seem to be a thriving industry).
John Polasek
>>
>>>

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#561713

On 3/11/2016 3:46 PM, Jackpol11@hotmail.com wrote:
> On Fri, 11 Mar 2016 11:47:57 -0600, Odd Bodkin <bodkinodd@gmail.com>
> wrote:
>
>> On 3/11/2016 11:18 AM, Jackpol11@hotmail.com wrote:
>>> On Fri, 11 Mar 2016 07:15:19 -0600, Odd Bodkin <bodkinodd@gmail.com>
>>> wrote:
>>>
>>>> On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
>>>>> For reference see:
>>>>> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
>>>>> I didn't say it was difficult-I said it was unspeakable. I worked out
>>>>> the units myself as you saw in the previous note so that's behind us.
>>>>> A bastard term like frequency to the -11/3 power needs someone to
>>>>> explain its relevance and its origin.
>>>>
>>>> The derivation of that equation can be found in the article referenced
>>>> in this paper. If you would bother to look that up, you might have your
>>>> questions answered.
>>>> L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
>>>> Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
>>>> This is the reason the reference is listed, so that you not think it was
>>>> made up from thin air.
>>> It's unavailable without a subscription.
>>
>> Whose problem is that supposed to be?
>> You have options:
>> 1. You purchase the article.
>> 2. You visit a university library that has purchased a subscription.
>>
>> One involves money, one involves legwork. Which of these are you willing
>> to invest?
 >>
> I think you're mistaking my volubility for fervor. I'm just looking
> for someone to explain the term frequency to the -11/3 power,
> augmented by 3/5  power, defining a unit of time, as the cofactor to
> c^3/G. But how do I know, out of 118 papers, that one of them will
> actually come across? I am not willing to gamble $35.

Then are you willing to invest the legwork, if you're not willing to pay 
for it?
This only means going to the library (which has a subscription) and 
sitting a while with the journals.

The reason I say that is that this is exactly what physicists do. They 
walk over to the library and they sit a while with the journals, 
reading. And if they don't find what they want with one article, they go 
to the next.

As for which article you should choose, you go to the article that is 
referenced at the point where your question is, which in this case was 
reference [11]. See how that works?

It sounds like you really don't want to either pay for it OR go to the 
library to read it. What you're hoping for is that, here where it is 
convenient, someone will take the trouble to elaborate on it to your 
satisfaction so that there's no work or investment required on your part.

> I evaluated the equation coming up with 620 µs for 32Msun. What does
> that time really represent?
> Let's hope they stay tuned for the next calamity.
> (Black holes seem to be a thriving industry).
> John Polasek
>>>
>>>>


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

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#561637 — was not aware physicists use phony irrational numbers to compensate for LIGO Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth?

On Friday, March 11, 2016 at 11:19:04 AM UTC-6, Jack...@hotmail.com wrote:
> On Fri, 11 Mar 2016 07:15:19 -0600, Odd Bodkin <bodkinodd@gmail.com>
> wrote:
> 
> >On 3/10/2016 9:21 PM, Jackpol11@hotmail.com wrote:
> >> For reference see:
> >> http://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102
> >> I didn't say it was difficult-I said it was unspeakable. I worked out
> >> the units myself as you saw in the previous note so that's behind us.
> >> A bastard term like frequency to the -11/3 power needs someone to
> >> explain its relevance and its origin.
> >
> >The derivation of that equation can be found in the article referenced 
> >in this paper. If you would bother to look that up, you might have your 
> >questions answered.
> >L. Blanchet, T. Damour, B. R. Iyer, C. M. Will, and A. G.
> >Wiseman, Phys. Rev. Lett. 74, 3515 (1995)
> >This is the reason the reference is listed, so that you not think it was 
> >made up from thin air.
> It's unavailable without a subscription.
> 
> >
> >> Or at least give it units so
> >> that it should be understood. for
> >>   Let L = wavelength or m/cycle = 1/F. Then we have
> >>       f^-11/3 = L^11/3 = cube root of (meters^11/cycle^11)
> >
> >I really don't think that's necessary. You're asking scientists to 
> >explain the units in a way that is understandable TO YOU. 
> No I am not. The units are where the physics is. Just making equations
> balance is not physics, if the equations are not the logical result of
> quantizing some physical sequence. f to the -11/3 is an example
> Of the 118 references in the paper, #11 was not available without a
> subscription. 
> 
> >> Things don't really improve by taking the 3/5 power of all that.
> >
> >I don't know what you mean by "improve". The theory is the theory.
> 
> I don't see a theory, I see an assertion. I see a believer, or no, a
> person who is unwilling to doubt. 
> There is another paper that gives a lot more information to the inside
> workings:
> http://arxiv.org/pdf/1602.03840v1.pdf
> It's hard to see how they identified the 2 masses as 30 and 32 solar
> masses, when the 1st notification was on earth at the time of the
> collision 1.3 Byrs later and had no idea in what direction to look
> (for 2 black holes).
> This is much like supernovae with a life several weeks, which appear
> at random, the best of which are identified as "no more than 5 days
> after the peak." Not one has been tracked from start to finish, so
> there is no functional form to describe the "light curve". it is
> necessary to "back-engineer" to the assumed peak. 
> (But easier than tracking those 2 black holes after they are
> demolished).
> 
> John Polasek

I am with John on this issue.

I suspect the archaic exponents come from a theory of multidimensions in which those are not warranted.

As Jack or John points out, Physics is in Quantum Mechanics of rational number like that of spin 1/2.

So if we start mickey mousing around with (.5)^11/3, the 11/3 power we have no longer rational numbers, and in the wide margin of 11/3 we end up with so much leeway error that we get the phony millisecond theory stretch of time in the LIGO interferometers.

AP

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#561989 — Re: was not aware physicists use phony irrational numbers to compensate for LIGO Re: why so many scientists and mathematicians opt for flash fame with wrong ideas rather than no fame but the truth?

So, do we have an independent auditor of "ping events" at LIGO, or, is the team the ones to report-- when and what pings are actually reported and other pings which would lend doubt on LIGO are just dismissed and trashed for they cast a dark shadow on the 14-15Sept2015 2 black holes?

Maybe LIGO had several such pings which by some independent body would have said all were 2 black holes.

AP

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