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(((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39

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  (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-26 20:34 -0700
    Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:33 -0700
      Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:44 -0700
        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:46 -0700
        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-27 08:53 -0700
          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 07:34 -0700
            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 07:46 -0700
              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 09:42 -0700
                Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:17 -0700
                  Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:25 -0700
                    Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:51 -0700
                      Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 10:54 -0700
                        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 19:29 -0700
                          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 19:59 -0700
                            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 10:43 -0700
                            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 10:44 -0700
                              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-29 11:03 -0700
                      Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-10-28 17:29 -0400
                        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 15:37 -0700
              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-28 09:51 -0700
          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-30 18:31 +0100
            Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 17:44 +0000
              Re: Something related to the Collision of Protons Odd Bodkin <bodkinodd@gmail.com> - 2015-10-30 14:11 -0500
                Re: Something related to the Collision of Protons Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-30 21:06 +0100
                  Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:47 +0000
                  Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:49 +0000
                  Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 20:54 +0000
                  Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 21:52 +0000
                  Re: Something related to the Collision of Protons Bobby Sifuentes <bobsif@academicinstitute.org> - 2015-10-30 21:59 +0000
                  Re: Something related to the Collision of Protons fuller.david@hotmail.com - 2015-10-30 22:20 -0700
            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 John Heath <heathjohn2@gmail.com> - 2015-10-30 16:10 -0700
              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-30 21:34 -0700
            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-30 21:32 -0700
              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 06:50 +0100
                Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Jack Weidner <jackwd@webportal.au> - 2015-10-31 11:20 +0000
                Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 05:48 -0700
                  Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 16:16 +0100
                    Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 08:32 -0700
                    Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 08:35 -0700
                      Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2015-10-31 17:18 +0100
                        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:30 -0700
                          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:38 -0700
                          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-10-31 13:42 -0400
                        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:58 -0700
                          Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 12:15 -0700
                            Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-11-04 07:58 -0800
                              Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-11-05 08:49 -0800
                                Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 kefischer <emoneyjoe@iglou.com> - 2015-11-05 12:08 -0500
                                Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Poutnik <Poutnik4NNTP@gmail.com> - 2015-11-05 19:23 +0100
                        Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 fuller.david@hotmail.com - 2015-10-31 12:11 -0700
                    Re: (((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 Hyperbolic.cc@outlook.com - 2015-10-31 09:59 -0700

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#368577

fuller.david@hotmail.com wrote:

> How much do the protons weigh in the LHC at 7Tev?
>  
> The energy of a proton is 7 TeV.

No, it is not.  That is the *collision* energy coming from at least *two* 
colliding protons.  You want to read page 3 and following of the 2009 
edition of the LHC Guide:

<http://cds.cern.ch/record/1165534/files/CERN-Brochure-2009-003-Eng.pdf#page-3>

> Via E = mc2 the mass is simply 7 TeV/c2 -
> and these are the units usually used.

["mc2" and "c2" are not proper notations.  If you cannot use Unicode -- 
"mc²" and "c²", respectively -- you have to find an ASCII-compatible way to 
express powers.  The usual way is to use the circumflex/caret character, so 
it would be "mc^2" and "c^2", respectively.]

As has been explained here many times before (including several times by 
me), E = m c² (E = m c^2) is only true for the energy at relative rest,
E = E(v = 0) = E(p = 0) =: E₀.

The full equation for energy, and thereby the mass–energy equivalence, that 
can be derived from the norm of the four-momentum is

  E = √((m c²)² + (p c)²)                                               (1)

 [E = sqrt((m c^2)^2 + (p c)^2)],

with momentum

  p = γ m v                                                             (2)

 [p = gamma m v],

and the Lorentz factor

  γ = 1∕√(1 – (v∕c)²)                                                   (3)

 [gamma = 1/sqrt(1 - (v/c)^2)],

therefore, applying (3) to (2),

  p = m v∕√(1 – (v∕c)²)                                                 (4)

 [p = m v/sqrt(1 - (v/c)^2)],

and finally, applying (4) to (1),

  E = m c³ √(1/(c² − v²))

 [E = m c^3 sqrt(1/(c^2 - v^2))],

assuming m > 0, c > 0, and v ≥ 0 (v >= 0).

As you can read in the LHC Guide, the protons in the LHC, when they reach 
their greatest relative velocity, move with 99.9999991 % of the speed of 
light in vacuum, or 0.999999991 c.  So it is very important not to ignore 
the part of their energy that comes from their relative motion.  In fact, 
that is why we need particle *accelerators* for achieving high particle 
energies, therefore the chance for producing heavy particles like the 
discovered Higgs boson, in the first place.
 
> 7 TeV/c^2 divided by the rest mass 0.938272029 GeV/c^2 gives us 7460.52
> times the rest mass

Ex falso quodlibet.
 

PointedEars
-- 
Q: What happens when electrons lose their energy?  
A: They get Bohr'ed.

(from: WolframAlpha)

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#368580 — Something related to the Collision of Protons

Thomas 'PointedEars' Lahn wrote:

>> The energy of a proton is 7 TeV.
> 
> No, it is not.  That is the *collision* energy coming from at least
> *two* colliding protons.

Well, would be quite impossible to collide a proton alone. What is the
cross-section of a proton?

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#368596 — Re: Something related to the Collision of Protons

On 10/30/2015 12:44 PM, Bobby Sifuentes wrote:
> Thomas 'PointedEars' Lahn wrote:
>
>>> The energy of a proton is 7 TeV.
>>
>> No, it is not.  That is the *collision* energy coming from at least
>> *two* colliding protons.
>
> Well, would be quite impossible to collide a proton alone. What is the
> cross-section of a proton?
>

Most of the time this is talked about as a "differential" cross section, 
where this means essentially the rate of scattering where the incoming 
particle is scattered by a particular angle.

This is a classical mechanics problem, and Rutherford worked it out when 
he was trying to figure out alpha-nucleus scattering.

https://en.wikipedia.org/wiki/Rutherford_scattering

The equation is the one directly *above* the section header "Details of 
calculating maximal nuclear size". For proton-proton scattering, Z=1.


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

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#368600 — Re: Something related to the Collision of Protons

Odd Bodkin wrote:

> On 10/30/2015 12:44 PM, Bobby Sifuentes wrote:
>> Thomas 'PointedEars' Lahn wrote:
>>>> The energy of a proton is 7 TeV.
>>> No, it is not.  That is the *collision* energy coming from at least
>>> *two* colliding protons.
>> Well, would be quite impossible to collide a proton alone. What is the
>> cross-section of a proton?
> 
> Most of the time this is talked about as a "differential" cross section,
> where this means essentially the rate of scattering where the incoming
> particle is scattered by a particular angle.
> 
> This is a classical mechanics problem, and Rutherford worked it out when
> he was trying to figure out alpha-nucleus scattering.
> 
> https://en.wikipedia.org/wiki/Rutherford_scattering
> 
> The equation is the one directly *above* the section header "Details of
> calculating maximal nuclear size". For proton-proton scattering, Z=1.

Congratulations, you actually managed to top the stupidity of the question 
of the ’nym-shifting troll with a wrong answer that has nothing to do with 
it.


PointedEars
-- 
“Science is empirical: knowing the answer means nothing;
 testing your knowledge means everything.”
   —Dr. Lawrence M. Krauss, theoretical physicist,
    in “A Universe from Nothing” (2009)

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#368604 — Re: Something related to the Collision of Protons

Thomas 'PointedEars' Lahn wrote:

>> The equation is the one directly *above* the section header "Details of
>> calculating maximal nuclear size". For proton-proton scattering, Z=1.
> 
> Congratulations, you actually managed to top the stupidity of the
> question of the ’nym-shifting troll with a wrong answer that has nothing
> to do with it.

You just said collision of *two* protons, and dare to call the others 
stupid and wrong?

PointEears:
"No, it is not.  That is the *collision* energy coming from at least
*two* colliding protons."

LOL.

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#368605 — Re: Something related to the Collision of Protons

Odd Bodkin wrote:

>> Congratulations, you actually managed to top the stupidity of the
>> question of the ’nym-shifting troll with a wrong answer that has
>> nothing to do with it.
> 
> Do tell http://arxiv.org/abs/1110.1395 http://arxiv.org/abs/1204.5689
> The total proton cross section is a combination of the inelastic and
> elastic cross sections.
> What would you suggest as a primer for elastic cross sections?

LOL, excellent answer Odd. Sad is wasted on Pointy.

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#368606 — Re: Something related to the Collision of Protons

Thomas 'PointedEars' Lahn wrote:

>> The equation is the one directly *above* the section header "Details of
>> calculating maximal nuclear size". For proton-proton scattering, Z=1.
> 
> Congratulations, you actually managed to top the stupidity of the
> question of the ’nym-shifting troll with a wrong answer that has nothing
> to do with it.

Hey Pointy, let me ask you in this way. Can the collision of one proton be 
directed towards something other than another proton? You are great, keep 
up the good work.

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#368608 — Re: Something related to the Collision of Protons

Odd Bodkin wrote:

>>> Do tell http://arxiv.org/abs/1110.1395 http://arxiv.org/abs/1204.5689
>>> The total proton cross section is a combination of the inelastic and
>>> elastic cross sections.
>>
>> If only you knew what you are talking about.  The papers you cite speak
>> of “proton-proton cross section” and “p+p reactions”.  The
>> ‘nym-shifting troll asked about the “cross-section of a [single]
>> proton”.  Granted, that is a
>> stupid question, but it does not mean that your answer has to be
>> equally stupid.
> 
> I certainly did not intend to answer a question about the cross-section
> of a single proton. If you're going to attempt to ridicule me for
> providing some steer from an obviously silly question, then allow me to
> ridicule you for being an anal-retentive, OCDish twat.

Why should that question be stupid? You are not as much stupid PointHead 
is. Or you both know nothing about cross-section ☭ Very funny indeed ♫♪♫♫

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#368609 — Re: Something related to the Collision of Protons

Thomas 'PointedEars' Lahn wrote:

>> Do tell http://arxiv.org/abs/1110.1395 http://arxiv.org/abs/1204.5689
>> The total proton cross section is a combination of the inelastic and
>> elastic cross sections.
> 
> If only you knew what you are talking about.  The papers you cite speak
> of “proton-proton cross section” and “p+p reactions”.  The ‘nym-shifting
> troll asked about the “cross-section of a [single] proton”.  Granted,
> that is a stupid question, but it does not mean that your answer has to
> be equally stupid.

The stupid must be you, pointhead. Cross-sections are IMAGINARY, pointhead, 
you don't need to cut. Say it to the other pointhead, so he can learn, 
cross-sections. ┌∩┐(◣_◢)┌∩┐

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#368648 — Re: Something related to the Collision of Protons

On Friday, October 30, 2015 at 3:06:53 PM UTC-5, Thomas 'PointedEars' Lahn wrote:
> Odd Bodkin wrote:
> 
> > On 10/30/2015 12:44 PM, Bobby Sifuentes wrote:
> >> Thomas 'PointedEars' Lahn wrote:
> >>>> The energy of a proton is 7 TeV.
> >>> No, it is not.  That is the *collision* energy coming from at least
> >>> *two* colliding protons.
> >> Well, would be quite impossible to collide a proton alone. What is the
> >> cross-section of a proton?
> > 
> > Most of the time this is talked about as a "differential" cross section,
> > where this means essentially the rate of scattering where the incoming
> > particle is scattered by a particular angle.
> > 
> > This is a classical mechanics problem, and Rutherford worked it out when
> > he was trying to figure out alpha-nucleus scattering.
> > 
> > https://en.wikipedia.org/wiki/Rutherford_scattering
> > 
> > The equation is the one directly *above* the section header "Details of
> > calculating maximal nuclear size". For proton-proton scattering, Z=1.
> 
> Congratulations, you actually managed to top the stupidity of the question 
> of the ’nym-shifting troll with a wrong answer that has nothing to do with 
> it.
> 
> 
> PointedEars
> -- 
> “Science is empirical: knowing the answer means nothing;
>  testing your knowledge means everything.”
>    —Dr. Lawrence M. Krauss, theoretical physicist,
>     in “A Universe from Nothing” (2009)


 E = √((m c²)² + (p c)²)                                               (1) 


Thanks for the Clarification, That is what I was Figuring ... 
I consider Space time ( the totality of ) to be "Kinetic Energy" .... so that fits RIGHT IN .. Thanks for the pointer. 

Mass exists Precisely because of the Vacuum's Impedance to velocity. 

The Mass = (376.73 / c) Times the Kinetic energy of the Vacuum = (376.73  c)*c^2 or simply (376.73 * c) which is 1 / (376.73 * the speed of light) = 8.85419518 × 10^-12 s / m 
Or the Reciprocal of the vacuum permittivity 

So is This ...[((((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 ] starting to Sink in YET ??? 

https://goo.gl/photos/FAr6s1oet3SUYvPn8

https://goo.gl/photos/kRF94hsbu6iaSfDA7

[E = sqrt((m c^2)^2 + (p c)^2)], 

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#368624

On Friday, October 30, 2015 at 1:31:40 PM UTC-4, Thomas 'PointedEars' Lahn wrote:
> fuller.david@hotmail.com wrote:
> 
> > How much do the protons weigh in the LHC at 7Tev?
> >  
> > The energy of a proton is 7 TeV.
> 
> No, it is not.  That is the *collision* energy coming from at least *two* 
> colliding protons.  You want to read page 3 and following of the 2009 
> edition of the LHC Guide:
> 
> <http://cds.cern.ch/record/1165534/files/CERN-Brochure-2009-003-Eng.pdf#page-3>
> 
> > Via E = mc2 the mass is simply 7 TeV/c2 -
> > and these are the units usually used.
> 
> ["mc2" and "c2" are not proper notations.  If you cannot use Unicode -- 
> "mc²" and "c²", respectively -- you have to find an ASCII-compatible way to 
> express powers.  The usual way is to use the circumflex/caret character, so 
> it would be "mc^2" and "c^2", respectively.]
> 
> As has been explained here many times before (including several times by 
> me), E = m c² (E = m c^2) is only true for the energy at relative rest,
> E = E(v = 0) = E(p = 0) =: E₀.
> 
> The full equation for energy, and thereby the mass–energy equivalence, that 
> can be derived from the norm of the four-momentum is
> 
>   E = √((m c²)² + (p c)²)                                               (1)
> 
>  [E = sqrt((m c^2)^2 + (p c)^2)],
> 
> with momentum
> 
>   p = γ m v                                                             (2)
> 
>  [p = gamma m v],
> 
> and the Lorentz factor
> 
>   γ = 1∕√(1 – (v∕c)²)                                                   (3)
> 
>  [gamma = 1/sqrt(1 - (v/c)^2)],
> 
> therefore, applying (3) to (2),
> 
>   p = m v∕√(1 – (v∕c)²)                                                 (4)
> 
>  [p = m v/sqrt(1 - (v/c)^2)],
> 
> and finally, applying (4) to (1),
> 
>   E = m c³ √(1/(c² − v²))
> 
>  [E = m c^3 sqrt(1/(c^2 - v^2))],
> 
> assuming m > 0, c > 0, and v ≥ 0 (v >= 0).
> 
> As you can read in the LHC Guide, the protons in the LHC, when they reach 
> their greatest relative velocity, move with 99.9999991 % of the speed of 
> light in vacuum, or 0.999999991 c.  So it is very important not to ignore 
> the part of their energy that comes from their relative motion.  In fact, 
> that is why we need particle *accelerators* for achieving high particle 
> energies, therefore the chance for producing heavy particles like the 
> discovered Higgs boson, in the first place.
>  
> > 7 TeV/c^2 divided by the rest mass 0.938272029 GeV/c^2 gives us 7460.52
> > times the rest mass
> 
> Ex falso quodlibet.
>  
> 
> PointedEars
> -- 
> Q: What happens when electrons lose their energy?  
> A: They get Bohr'ed.
> 
> (from: WolframAlpha)

You know the meaning of mc2 and c2 so why are you knit picking Fuller on keyboard fonts and the " standard format " if you knew what his meaning was? This is off subject to the question at hand yes / no ?

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#368641

On Friday, October 30, 2015 at 6:10:42 PM UTC-5, John Heath wrote:
> On Friday, October 30, 2015 at 1:31:40 PM UTC-4, Thomas 'PointedEars' Lahn wrote:
> > fuller.david@hotmail.com wrote:
> > 
> > > How much do the protons weigh in the LHC at 7Tev?
> > >  
> > > The energy of a proton is 7 TeV.
> > 
> > No, it is not.  That is the *collision* energy coming from at least *two* 
> > colliding protons.  You want to read page 3 and following of the 2009 
> > edition of the LHC Guide:
> > 
> > <http://cds.cern.ch/record/1165534/files/CERN-Brochure-2009-003-Eng.pdf#page-3>
> > 
> > > Via E = mc2 the mass is simply 7 TeV/c2 -
> > > and these are the units usually used.
> > 
> > ["mc2" and "c2" are not proper notations.  If you cannot use Unicode -- 
> > "mc²" and "c²", respectively -- you have to find an ASCII-compatible way to 
> > express powers.  The usual way is to use the circumflex/caret character, so 
> > it would be "mc^2" and "c^2", respectively.]
> > 
> > As has been explained here many times before (including several times by 
> > me), E = m c² (E = m c^2) is only true for the energy at relative rest,
> > E = E(v = 0) = E(p = 0) =: E₀.
> > 
> > The full equation for energy, and thereby the mass–energy equivalence, that 
> > can be derived from the norm of the four-momentum is
> > 
> >   E = √((m c²)² + (p c)²)                                               (1)
> > 
> >  [E = sqrt((m c^2)^2 + (p c)^2)],
> > 
> > with momentum
> > 
> >   p = γ m v                                                             (2)
> > 
> >  [p = gamma m v],
> > 
> > and the Lorentz factor
> > 
> >   γ = 1∕√(1 – (v∕c)²)                                                   (3)
> > 
> >  [gamma = 1/sqrt(1 - (v/c)^2)],
> > 
> > therefore, applying (3) to (2),
> > 
> >   p = m v∕√(1 – (v∕c)²)                                                 (4)
> > 
> >  [p = m v/sqrt(1 - (v/c)^2)],
> > 
> > and finally, applying (4) to (1),
> > 
> >   E = m c³ √(1/(c² − v²))
> > 
> >  [E = m c^3 sqrt(1/(c^2 - v^2))],
> > 
> > assuming m > 0, c > 0, and v ≥ 0 (v >= 0).
> > 
> > As you can read in the LHC Guide, the protons in the LHC, when they reach 
> > their greatest relative velocity, move with 99.9999991 % of the speed of 
> > light in vacuum, or 0.999999991 c.  So it is very important not to ignore 
> > the part of their energy that comes from their relative motion.  In fact, 
> > that is why we need particle *accelerators* for achieving high particle 
> > energies, therefore the chance for producing heavy particles like the 
> > discovered Higgs boson, in the first place.
> >  
> > > 7 TeV/c^2 divided by the rest mass 0.938272029 GeV/c^2 gives us 7460.52
> > > times the rest mass
> > 
> > Ex falso quodlibet.
> >  
> > 
> > PointedEars
> > -- 
> > Q: What happens when electrons lose their energy?  
> > A: They get Bohr'ed.
> > 
> > (from: WolframAlpha)
> 
> You know the meaning of mc2 and c2 so why are you knit picking Fuller on keyboard fonts and the " standard format " if you knew what his meaning was? This is off subject to the question at hand yes / no ?

Thanks for Having My Back John ....

Look at this if you Get Bohred ....

https://plus.google.com/107909633052051852299/posts/9ujLRBLkfUH

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#368640

On Friday, October 30, 2015 at 12:31:40 PM UTC-5, Thomas 'PointedEars' Lahn wrote:
> fuller.david@hotmail.com wrote:
> 
> > How much do the protons weigh in the LHC at 7Tev?
> >  
> > The energy of a proton is 7 TeV.
> 
> No, it is not.  That is the *collision* energy coming from at least *two* 
> colliding protons.  You want to read page 3 and following of the 2009 
> edition of the LHC Guide:
> 
> <http://cds.cern.ch/record/1165534/files/CERN-Brochure-2009-003-Eng.pdf#page-3>
> 
> > Via E = mc2 the mass is simply 7 TeV/c2 -
> > and these are the units usually used.
> 
> ["mc2" and "c2" are not proper notations.  If you cannot use Unicode -- 

Yeah, Sorry ..That was just Garbage ["mc2" and "c2"] I copied and pasted from the CERN Site. normally I correct it. Big Difference between 4pi and 4^pi

> "mc²" and "c²", respectively -- you have to find an ASCII-compatible way to 
> express powers.  The usual way is to use the circumflex/caret character, so 
> it would be "mc^2" and "c^2", respectively.]
> 
> As has been explained here many times before (including several times by 
> me), E = m c² (E = m c^2) is only true for the energy at relative rest,
> E = E(v = 0) = E(p = 0) =: E₀.
> 
> The full equation for energy, and thereby the mass–energy equivalence, that 
> can be derived from the norm of the four-momentum is
> 
>   E = √((m c²)² + (p c)²)                                               (1)

Thanks for the Clarification, That is what I was Figuring ...
I consider Space time ( the totality of)to be "Kinetic Energy" .... so that fits RIGHT IN .. Thanks for the pointer.

Mass exists Precisely because of the Vacuum's Impedance to velocity.

The Mass = (376.73 / c) 
Times the Kinetic energy of the Vacuum = (376.73  c)*c^2 or simply (376.73 * c)
which is 1 / (376.73 * the speed of light) = 8.85419518 × 10^-12 s / m
Or the Reciprocal of the vacuum permittivity 

So is This ...[((((4 * pi * (10^(-7)))^2) / 2)^pi = 9.5176229e-39 ] starting to Sink in YET ???


> 
>  [E = sqrt((m c^2)^2 + (p c)^2)],
> 
> with momentum
> 
>   p = γ m v                                                             (2)
> 
>  [p = gamma m v],
> 
> and the Lorentz factor
> 
>   γ = 1∕√(1 – (v∕c)²)                                                   (3)
> 
>  [gamma = 1/sqrt(1 - (v/c)^2)],
> 
> therefore, applying (3) to (2),
> 
>   p = m v∕√(1 – (v∕c)²)                                                 (4)
> 
>  [p = m v/sqrt(1 - (v/c)^2)],
> 
> and finally, applying (4) to (1),
> 
>   E = m c³ √(1/(c² − v²))
> 
>  [E = m c^3 sqrt(1/(c^2 - v^2))],
> 
> assuming m > 0, c > 0, and v ≥ 0 (v >= 0).

Yep yep All yer Figuring looks Right on the Money.... Thanks ...
> 
> As you can read in the LHC Guide, the protons in the LHC, when they reach 
> their greatest relative velocity, move with 99.9999991 % of the speed of 
> light in vacuum, or 0.999999991 c.  So it is very important not to ignore 
> the part of their energy that comes from their relative motion.  In fact, 
> that is why we need particle *accelerators* for achieving high particle 
> energies, therefore the chance for producing heavy particles like the 
> discovered Higgs boson, in the first place.
>  
> > 7 TeV/c^2 divided by the rest mass 0.938272029 GeV/c^2 gives us 7460.52
> > times the rest mass
> 
> Ex falso quodlibet.

Fancy Latin ... No thanks, More Clutter. Nope Nope ....

>  
> 
> PointedEars
> -- 
> Q: What happens when electrons lose their energy?  
> A: They get Bohr'ed.
> 
> (from: WolframAlpha)

Thanks for the Assistance !!!

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#368651

fuller.david@hotmail.com wrote:

> On Friday, October 30, 2015 at 12:31:40 PM UTC-5, Thomas 'PointedEars'
> Lahn wrote:
>> The full equation for energy, and thereby the mass–energy equivalence,
>> that can be derived from the norm of the four-momentum is
>> 
>>   E = √((m c²)² + (p c)²)                                              
>>   (1)
> 
> Thanks for the Clarification, That is what I was Figuring ...
> I consider Space time ( the totality of)to be "Kinetic Energy"

You want to reconsider.  Spacetime is a mathematical framework for 
describing reality instead.  E is the energy of an object in spacetime — 
which does not need to move relative to the observer; p can be 0 and the 
object still has energy —, not the energy of spacetime itself.  In fact,
the total energy of the universe is very likely 0.

The relativistic kinetic energy of a rigid body is instead

  E_k = m γ c² – m c².

Note how with γ = 1∕√(1 – (v∕c)²) the kinetic energy of a rigid body *also* 
depends on the frame of reference as v is the speed of the object relative 
to the stationary observer, and that it becomes zero if m γ c² = m c²; that 
happens precisely when γ = 1 which requires that v = 0 (no relative motion).

See also:

<http://www.britannica.com/topic/Albert-Einstein-on-Space-Time-1987141>
<https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies>

> […]
> Mass exists Precisely because of the Vacuum's Impedance to velocity.

That is such a nonsense, it is not even wrong.


PointedEars
-- 
Q: Why is electricity so dangerous?  
A: It doesn't conduct itself.

(from: WolframAlpha)

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#368669

Thomas 'PointedEars' Lahn wrote:

> You want to reconsider.  Spacetime is a mathematical framework for
> describing reality instead.  E is the energy of an object in spacetime —
> which does not need to move relative to the observer; p can be 0 and the
> object still has energy —, not the energy of spacetime itself.  In fact,
> the total energy of the universe is very likely 0.

Idiot.

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#368682

On Saturday, October 31, 2015 at 12:50:37 AM UTC-5, Thomas 'PointedEars' Lahn wrote:
> fuller.david@hotmail.com wrote:
> 
> > On Friday, October 30, 2015 at 12:31:40 PM UTC-5, Thomas 'PointedEars'
> > Lahn wrote:
> >> The full equation for energy, and thereby the mass–energy equivalence,
> >> that can be derived from the norm of the four-momentum is
> >> 
> >>   E = √((m c²)² + (p c)²)                                              
> >>   (1)
> > 
> > Thanks for the Clarification, That is what I was Figuring ...
> > I consider Space time ( the totality of)to be "Kinetic Energy"
> 
> You want to reconsider.  Spacetime is a mathematical framework for 
> describing reality instead.  E is the energy of an object in spacetime — 
> which does not need to move relative to the observer; p can be 0 and the 
> object still has energy —, not the energy of spacetime itself.  In fact,
> the total energy of the universe is very likely 0.
> 
> The relativistic kinetic energy of a rigid body is instead
> 
>   E_k = m γ c² – m c².
> 
> Note how with γ = 1∕√(1 – (v∕c)²) the kinetic energy of a rigid body *also* 
> depends on the frame of reference as v is the speed of the object relative 
> to the stationary observer, and that it becomes zero if m γ c² = m c²; that 
> happens precisely when γ = 1 which requires that v = 0 (no relative motion).
> 
> See also:
> 
> <http://www.britannica.com/topic/Albert-Einstein-on-Space-Time-1987141>
> <https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies>
> 
> > […]
> > Mass exists Precisely because of the Vacuum's Impedance to velocity.
> 
> That is such a nonsense, it is not even wrong

No, that is you just performing the "Hand Waving Maneuver"
Do it this way then

sqrt(The pi root of 10^-39 / 2) = vacuum permeability approx.

The vacuum impedes velocity


> 
> 
> PointedEars
> -- 
> Q: Why is electricity so dangerous?  
> A: It doesn't conduct itself.
> 
> (from: WolframAlpha)

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#368701

fuller.david@hotmail.com wrote:

> On Saturday, October 31, 2015 at 12:50:37 AM UTC-5, Thomas 'PointedEars'
> Lahn wrote:
>> fuller.david@hotmail.com wrote:
>> > […]
>> > Mass exists Precisely because of the Vacuum's Impedance to velocity.
>> 
>> That is such a nonsense, it is not even wrong
> 
> No, that is you just performing the "Hand Waving Maneuver"
> Do it this way then
> 
> sqrt(The pi root of 10^-39 / 2) = vacuum permeability approx.

Fascinating.  But you probably did not know that years ago a reknowned Dutch 
astronomer, Cornelis de Jager, found out that many of the fundamental 
physical constants, and even astronomical measures, are actually provided by 
Dutch roadsters (a kind of city bicycle).

I could not find the original article [1] online, so I thought I should give 
you the privilege of learning about the most important parts of this amazing 
discovery and new school of thought, called Cyclosophy (in Dutch: 
Velosofie), from me, right here (taken from German translations that I have 
found):

If one defines the quantities, as carefully measured on the aforementioned 
vehicle,

  P – pedal travel
  W – front wheel diameter 
  L – lamp diameter
  B – bell diameter

then it turns out that

          P² √(L B) = 1823 ≈ m_p∕m_e (proton mass∕electron mass)

              W²∕P⁴ = 1∕137 ≈ α (fine structure constant)

  1∕P⁵ 3 ∛(L∕(W B)) = 6.67 × 10⁻⁸ ≈ G (gravitational constant in ft³∕slugs²)

            √P ∛B/L = 1.496 ≈ 1 au (in units of 10⁸ km)
 
and finally (if you are not sitting already, you better sit down before you 
read this), taking all measurements into account:

       W^π P² ∛L B⁵ = 2.99 × 10⁵ ≈ c (speed of light in vacuum if km∕s = 1)!

Isn’t that amazing?  (<rot13>Ab.  Qr Wntre fubjf gung lbh pna pbzovar nal 
sbhe ahzoref guvf jnl – jvgu fdhner ebbgf, phovp ebbgf, naq cbjref 
rfcrpvnyyl bs π – qrevir gubfr pbafgnagf naq znal zber vzcbegnag ahzoref.  
Gurer ner znal jnlf gb pbzovar gurz, fb gung gurer vf n tbbq punapr gung lbh 
trg nccebkvzngryl gur ahzoref lbh ner ybbxvat sbe.</rot13>)

Learn more:

[1] De Jager, C. (1990). Velosofie: Rekenen aan de Grote Piramide en m'n
    fiets (“Calculating on the Large Pyramid and on my bicycle”), Skepter,
    3(4), pp. 13–15.

    De Jager, C. (1992). Adventures in science and cyclosophy.
    Skeptical Inquirer 16, pp. 167–172.
 
> The vacuum impedes velocity

Vacuum is a medium that is devoid of matter that could hinder the motion of 
particles traveling through it (by them colliding).  It is therefore the 
best medium to enable motion, not to hinder it.  For that reason, for 
example the speed of light is greatest in a vacuum.

F'up2 sci.physics


PointedEars
-- 
Two neutrinos go through a bar ...

(from: WolframAlpha)

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#368704

On Saturday, October 31, 2015 at 10:16:22 AM UTC-5, Thomas 'PointedEars' Lahn wrote:
> fuller.david@hotmail.com wrote:
> 
> > On Saturday, October 31, 2015 at 12:50:37 AM UTC-5, Thomas 'PointedEars'
> > Lahn wrote:
> >> fuller.david@hotmail.com wrote:
> >> > […]
> >> > Mass exists Precisely because of the Vacuum's Impedance to velocity.
> >> 
> >> That is such a nonsense, it is not even wrong
> > 
> > No, that is you just performing the "Hand Waving Maneuver"
> > Do it this way then
> > 
> > sqrt(The pi root of 10^-39 / 2) = vacuum permeability approx.
> 
> Fascinating.  But you probably did not know that years ago a reknowned Dutch 
> astronomer, Cornelis de Jager, found out that many of the fundamental 
> physical constants, and even astronomical measures, are actually provided by 
> Dutch roadsters (a kind of city bicycle).
> 
> I could not find the original article [1] online, so I thought I should give 
> you the privilege of learning about the most important parts of this amazing 
> discovery and new school of thought, called Cyclosophy (in Dutch: 
> Velosofie), from me, right here (taken from German translations that I have 
> found):
> 
> If one defines the quantities, as carefully measured on the aforementioned 
> vehicle,
> 
>   P – pedal travel
>   W – front wheel diameter 
>   L – lamp diameter
>   B – bell diameter
> 
> then it turns out that
> 
>           P² √(L B) = 1823 ≈ m_p∕m_e (proton mass∕electron mass)
> 
>               W²∕P⁴ = 1∕137 ≈ α (fine structure constant)
> 
>   1∕P⁵ 3 ∛(L∕(W B)) = 6.67 × 10⁻⁸ ≈ G (gravitational constant in ft³∕slugs²)
> 
but that just works out to 

1/(c*50) = G  .. It is Not Exact 

1 / (the speed of light * 50) = 6.6712819 × 10^-11 s / m

The "WHY" is completely IGNORED

WHY = https://goo.gl/photos/xQvTEFVJPSHhwNEV7 

https://goo.gl/photos/1wLBgxBT8ySVAJoM6 


Think (Surface area m^2 of a Sphere - 1^2) ^2 ... 
1/(1^2) now = 1- (1 / 24.1327412287) = 0.95856252 

1 - 0.95856252 = 0.04143748 

137 + 1/((4pi)^2 - (4pi-1)^2) = 137.041437481 


pi / ((11 * 2) / 7) = 0.9995976625 






>             √P ∛B/L = 1.496 ≈ 1 au (in units of 10⁸ km)
>  
> and finally (if you are not sitting already, you better sit down before you 
> read this), taking all measurements into account:
> 
>        W^π P² ∛L B⁵ = 2.99 × 10⁵ ≈ c (speed of light in vacuum if km∕s = 1)!
> 
> Isn’t that amazing?  (<rot13>Ab.  Qr Wntre fubjf gung lbh pna pbzovar nal 
> sbhe ahzoref guvf jnl – jvgu fdhner ebbgf, phovp ebbgf, naq cbjref 
> rfcrpvnyyl bs π – qrevir gubfr pbafgnagf naq znal zber vzcbegnag ahzoref.  
> Gurer ner znal jnlf gb pbzovar gurz, fb gung gurer vf n tbbq punapr gung lbh 
> trg nccebkvzngryl gur ahzoref lbh ner ybbxvat sbe.</rot13>)
> 
> Learn more:
> 
> [1] De Jager, C. (1990). Velosofie: Rekenen aan de Grote Piramide en m'n
>     fiets (“Calculating on the Large Pyramid and on my bicycle”), Skepter,
>     3(4), pp. 13–15.
> 
>     De Jager, C. (1992). Adventures in science and cyclosophy.
>     Skeptical Inquirer 16, pp. 167–172.
>  
> > The vacuum impedes velocity
> 
> Vacuum is a medium that is devoid of matter that could hinder the motion of 
> particles traveling through it (by them colliding).  It is therefore the 
> best medium to enable motion, not to hinder it.  For that reason, for 
> example the speed of light is greatest in a vacuum.
> 
> F'up2 sci.physics
> 
> 
> PointedEars
> -- 
> Two neutrinos go through a bar ...
> 
> (from: WolframAlpha)

[toc] | [prev] | [next] | [standalone]

#368706

On Saturday, October 31, 2015 at 10:16:22 AM UTC-5, Thomas 'PointedEars' Lahn wrote:
> fuller.david@hotmail.com wrote:
> 
> > On Saturday, October 31, 2015 at 12:50:37 AM UTC-5, Thomas 'PointedEars'
> > Lahn wrote:
> >> fuller.david@hotmail.com wrote:
> >> > […]
> >> > Mass exists Precisely because of the Vacuum's Impedance to velocity.
> >> 
> >> That is such a nonsense, it is not even wrong
> > 
> > No, that is you just performing the "Hand Waving Maneuver"
> > Do it this way then
> > 
> > sqrt(The pi root of 10^-39 / 2) = vacuum permeability approx.
> 
> Fascinating.  But you probably did not know that years ago a reknowned Dutch 
> astronomer, Cornelis de Jager, found out that many of the fundamental 
> physical constants, and even astronomical measures, are actually provided by 
> Dutch roadsters (a kind of city bicycle).
> 
> I could not find the original article [1] online, so I thought I should give 
> you the privilege of learning about the most important parts of this amazing 
> discovery and new school of thought, called Cyclosophy (in Dutch: 
> Velosofie), from me, right here (taken from German translations that I have 
> found):
> 
> If one defines the quantities, as carefully measured on the aforementioned 
> vehicle,
> 
>   P – pedal travel
>   W – front wheel diameter 
>   L – lamp diameter
>   B – bell diameter
> 
> then it turns out that
> 
>           P² √(L B) = 1823 ≈ m_p∕m_e (proton mass∕electron mass)
> 
>               W²∕P⁴ = 1∕137 ≈ α (fine structure constant)
> 
>   1∕P⁵ 3 ∛(L∕(W B)) = 6.67 × 10⁻⁸ ≈ G (gravitational constant in ft³∕slugs²)
> 
>             √P ∛B/L = 1.496 ≈ 1 au (in units of 10⁸ km)
>  
> and finally (if you are not sitting already, you better sit down before you 
> read this), taking all measurements into account:
> 
>        W^π P² ∛L B⁵ = 2.99 × 10⁵ ≈ c (speed of light in vacuum if km∕s = 1)!

I simplified THAT mess for you ......

Magnetic permeability = (4pi * 10^-7)

(4pi * 10^-7)/ ((11^2 + 4^2)*(11/4) = 3.33546665e-9 Hz

1/((4pi * 10^-7)/ ((11^2 + 4^2)*(11/4) )= 299808124.049




> 
> Isn’t that amazing?  (<rot13>Ab.  Qr Wntre fubjf gung lbh pna pbzovar nal 
> sbhe ahzoref guvf jnl – jvgu fdhner ebbgf, phovp ebbgf, naq cbjref 
> rfcrpvnyyl bs π – qrevir gubfr pbafgnagf naq znal zber vzcbegnag ahzoref.  
> Gurer ner znal jnlf gb pbzovar gurz, fb gung gurer vf n tbbq punapr gung lbh 
> trg nccebkvzngryl gur ahzoref lbh ner ybbxvat sbe.</rot13>)
> 
> Learn more:
> 
> [1] De Jager, C. (1990). Velosofie: Rekenen aan de Grote Piramide en m'n
>     fiets (“Calculating on the Large Pyramid and on my bicycle”), Skepter,
>     3(4), pp. 13–15.
> 
>     De Jager, C. (1992). Adventures in science and cyclosophy.
>     Skeptical Inquirer 16, pp. 167–172.
>  
> > The vacuum impedes velocity
> 
> Vacuum is a medium that is devoid of matter that could hinder the motion of 
> particles traveling through it (by them colliding).  It is therefore the 
> best medium to enable motion, not to hinder it.  For that reason, for 
> example the speed of light is greatest in a vacuum.
> 
> F'up2 sci.physics
> 
> 
> PointedEars
> -- 
> Two neutrinos go through a bar ...
> 
> (from: WolframAlpha)

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#368711

fuller.david@hotmail.com wrote:

> On Saturday, October 31, 2015 at 10:16:22 AM UTC-5, Thomas 'PointedEars'
> Lahn wrote:
>> If one defines the quantities, as carefully measured on the
>> aforementioned vehicle,
>> 
>>   P – pedal travel
>>   W – front wheel diameter
>>   L – lamp diameter
>>   B – bell diameter
>> 
>> then it turns out that
>> […]
>>        W^π P² ∛L B⁵ = 2.99 × 10⁵ ≈ c (speed of light in vacuum if km∕s =
>>        1)!
> 
> I simplified THAT mess for you ......

There is no mess to simplify, you poor lunatic.  It is nonsense intended to 
be that; actually, it is sarcasm about crazy people like you who arbitrarily 
put in numbers (like, “who cares about dimensions?”) and actually assign 
meaning to the inevitable outcome of these plays with numbers.
 
> Magnetic permeability = (4pi * 10^-7)

Do you even read what you are replying to?  B is _not_ meant to be the 
magnetic field here; it is the bell diameter, the diameter of a bicycle’s 
bell – bell as in ringing, not in Bell Labs.  So the magnetic permeability 
does not even feature here.

And I can’t believe that you are taking this seriously.  Get some 
professional help, will you, *please*?
 

PointedEars
-- 
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

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