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A new Momentum Theory for the Gravitation

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  A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-23 23:13 +0200
    Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-24 10:57 +0200
      Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-24 18:07 +0200
        Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 05:11 +0200
          Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 09:42 +0200
    A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-24 09:48 -0700
      A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-24 09:53 -0700
        A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-24 09:59 -0700
        Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-26 11:07 -0700
          Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-26 20:52 +0200
            Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-26 12:15 -0700
              Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-26 15:49 -0700
                Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-26 16:07 -0700
                  Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-27 00:37 -0700
                    Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-27 07:33 -0700
                      Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-27 07:55 -0700
                      Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-27 08:11 -0700
                        Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-27 08:35 -0700
                    Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-27 08:48 -0700
                      Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-27 10:07 -0700
                        Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-27 13:09 -0700
                          Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-28 07:27 -0700
                            Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-28 08:11 -0700
                              Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-28 13:37 -0700
                                Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-28 15:27 -0700
                                  Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-28 17:01 -0700
                                    Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-28 22:08 -0700
                                      Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-29 08:18 -0700
                                        Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-29 08:38 -0700
                                          Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-29 08:50 -0700
                                            Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-29 09:10 -0700
                                              Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-29 09:30 -0700
                                        Re: A new Momentum Theory for the Gravitation David Waite <waitedavid1618@yahoo.com> - 2016-04-30 09:03 -0700
            Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-26 15:37 -0700
            Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-04-26 16:00 -0700
            Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 10:45 +0200
              Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-05-03 10:57 -0700
                Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-05-03 20:13 +0200
              Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-05-03 11:07 -0700
                Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-05-03 20:14 +0200
                  Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-05-03 11:47 -0700
                    Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-05-03 21:46 +0200
                      Re: A new Momentum Theory for the Gravitation David Fuller <fuller.david@hotmail.com> - 2016-05-03 14:32 -0700
    Re: A new Momentum Theory for the Gravitation John Heath <heathjohn2@gmail.com> - 2016-04-26 17:12 -0700
      Re: A new Momentum Theory for the Gravitation "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2016-04-26 21:15 -0700
        Re: A new Momentum Theory for the Gravitation "Ross A. Finlayson" <ross.finlayson@gmail.com> - 2016-04-26 21:32 -0700
      Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-29 19:42 +0200
        Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 05:50 +0200
          Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-30 17:01 +0200
            Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 18:13 +0200
              Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-30 18:44 +0200
                Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-05-01 04:12 +0200
        Re: A new Momentum Theory for the Gravitation Fritz Lasinger <las8@web.de> - 2016-04-30 17:31 +0200
          Re: A new Momentum Theory for the Gravitation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-04-30 18:37 +0200

Page 3 of 3 — ← Prev page 1 2 [3]

#382722

On Tuesday, May 3, 2016 at 1:14:42 PM UTC-5, Thomas 'PointedEars' Lahn wrote:
> David Fuller wrote:
> 
> > On Saturday, April 30, 2016 at 3:45:09 AM UTC-5, Thomas 'PointedEars' Lahn
> > wrote:
> >> Fritz Lasinger wrote:
> >> > Hy David, I'm not sure, what you want to say?
> >> 
> >> David Fuller is, or at least he presents himself here as, a numerologist
> >> who lives under the delusion that what he is doing would resemble
> >> science. For your own sanity, it is better to ignore him.
> >> 
> >> [full quote]
> > 
> > von Klitzing constant RK = h/e2 = 25812.807557(18) Ω.
> > 
> > 1 / ((3 * (10^8) * 4 * 3.12500 * (10^(-7))) / (135 * 375)) =  135
> > […]
> > Space Time is Kinetic Energy
> 
> Q.E.D.
> 
> -- 
> PointedEars
> 
> Twitter: @PointedEars2
> Please do not cc me. / Bitte keine Kopien per E-Mail.


1 / ((3 * (10^8) * 4 * 3.12500 * (10^(-7))) / (135 * 375)) =  135 

1 / ((c * 4 * pi * (10^(-7))) / (2 * 25812.80755757)) = 137.03600 

1 / ((c * 4 * pi * (10^(-7))) / (376.730313 * 137.035999)) = 137.03600 

von Klitzing constant RK = (376.730313 * 137.03599984533232928352118030916) /2 

(375 * 135)^0.25 = 15 * 10^9 

 1 / (15 * (10^9)) = 6.6666667e-11 

(376.730313 * 137.03599984533232928352118030916)^0.25 = 15.0735765153 

1/(1.50735765153 * 10^10) = 6.6341256e-11 

Space Time is Kinetic Energy 

 1507/11 = 137 

The Geometry is Proper, pi is just a little Contracted to 3.125

LC = 0.99471839432434584855552352107821

v of c * 0.10264168740231828

the speed of light * 0.10264168740231828 = 30 771 203.8 m / s

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#382729

David Fuller wrote:

> 1 / ((3 * (10^8) * 4 * 3.12500 * (10^(-7))) / (135 * 375)) =  135
> 
> 1 / ((c * 4 * pi * (10^(-7))) / (2 * 25812.80755757)) = 137.03600
> 
> 1 / ((c * 4 * pi * (10^(-7))) / (376.730313 * 137.035999)) = 137.03600
> 
> von Klitzing constant RK = (376.730313 *
> 137.03599984533232928352118030916) /2
> 
> (375 * 135)^0.25 = 15 * 10^9
> 
>  1 / (15 * (10^9)) = 6.6666667e-11
> 
> (376.730313 * 137.03599984533232928352118030916)^0.25 = 15.0735765153
> 
> 1/(1.50735765153 * 10^10) = 6.6341256e-11
> 
> Space Time is Kinetic Energy
> 
>  1507/11 = 137
> 
> The Geometry is Proper, pi is just a little Contracted to 3.125
> 
> LC = 0.99471839432434584855552352107821
> 
> v of c * 0.10264168740231828
> 
> the speed of light * 0.10264168740231828 = 30 771 203.8 m / s

Witnessing this mindbogglingly stupid braindump of a deeply
troubled mind, suddenly even Vogon poetry appears bearable.

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#382741

Witnessing this mindbogglingly stupid braindump of a deeply 
troubled mind, suddenly even Vogon poetry appears bearable. 
- hide quoted text -

-- 
PointedEars 

Twitter: @PointedEars2 
Eat a dick Fucktard

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#382317

On Saturday, April 23, 2016 at 5:13:33 PM UTC-4, Fritz Lasinger wrote:
> Hello, I would like to present and discuss my new Momentum Theory for 
> the Gravitation. I'm very interested in your opinions!
> 
> 
> It will be shown, that gravitons only by momentum exchange exert a 
> force, which is inversely proportional to their distance. If the energy, 
> which is stored in the mass of the gravitons, corresponds to the 
> potential energy of the attracting masses, the
> gravitons cause exactly the gravitational force by the momentum 
> exchange. Thereby the graviton mass becomes negative. Finally the 
> assumption, that the sum of all positiv masses and negative graviton 
> masses in the universe is zero, leads to a plausible positive mass of 
> the universe of Mu ≈ 1.0·10^54 kg.
> 
> 
> Introduction
> 
> In the following three hypotheses will be presented and it will be 
> tried, to develop out of them the law of gravitation and the mass of the 
> universe. The three hypotheses are:
> 
> 1. A so called graviton exists with the mass mG, which is moving at the
> speed of light c between two masses m1 und m2 back and forth.
> 
> 2. The potential energy of the gravitation is found in the mass of the 
> gravitons.
> 
> 3. The sum of all mass is zero.
> 
> 
> Hypothesis 1: A so called graviton exists with the mass mG, which is 
> moving at the speed of light c between two masses m1 and m2 back and forth.
> 
> An observer is moving with the same speed as mass m1. The mass m2 shall
> have the distance x to the mass m1 and moves away from it with the speed 
> v. The observer sees the graviton, moving toward the mass m2 with the 
> relative momentum P = mG·(c−v). It fulfils a fully elastic impact and 
> after that it moves backwards away from the mass m2 with the same 
> relative momentum P = (mG+∆mG)·(c+v). Since the graviton cannot get 
> slower, its mass is changing by ∆mG. The amount of the momentum has 
> remained unchanged:
> mG·(c−v) = (mG+∆mG)·(c+v) respectively ∆mG = mG·((c−v)/(c+v)-1)
> 
> ∆mG/mG = −2v/(c+v)     (1)
> 
> The graviton transfers the following momentum to the mass m2:
> ∆P = mG·(c−v)+(mG+∆mG)·(c+v)
> 
> ∆P = 2·mG·c+∆mG·(c+v)     (2)
> 
> The fl ight time back and forth follows from ∆t/2·(c−v) = x
> 
> ∆t = 2x/(c−v)     (3)
> 
> Meanwhile the distance between m1 and m2 increases by
> ∆x = v·∆t = 2x·v/(c−v).
> 
> ∆x/x = 2v/(c-v)     (4)
> 
> (1)/(4) gives the change of the graviton mass by the movement ∆x of mass m2:
> (∆mG/mG)·(2v/(c-v)) = -(∆x/x)·2v/(c+v) respectively
> 
> ∆mG/mG = -((c-v)/(c+v))·(∆x/x)     (5)
> 
> To be able to integrate eq. 5, it is assumed v<<c for simpli cation. 
> Thereby follows dmG/mg = -dx/x -> Integral(dmG/mG)=-Integral(dx/x) 
> respectively
> 
> mG = mG0·x0/x     (6)
> 
> The average momentum fl ow, i. e. the average force, exerting on mass m2,
> results from the transfered momentum (eq. 2), devided by the time (eq. 
> 3) F = ∆P/∆t = ((2mG·c+∆mG·(c+v))/(2x))·(c−v). The equation is simpli ed 
> by v<<c. Thereby follows F = mG·c²/x respectively
> 
> F·x = mG·c^2     (7)
> 
> Insertion of equation (6) leads to
> 
> F = mG0·c^2·x0/x^2     (8)
> 
> With F0 = mG0·c^2/x0 follows F = F0·x0^2/x^2. Hence the graviton exerts 
> an average force F ∼ 1/x^2 to the masses m1 and m2, when it is assumed 
> v<<c for simpli cation.
> 
> Eq. 7 suggests to claim the next hypothesis:
> 
> 
> Hypothesis 2: The potential energy of the gravitation is
> found in the mass of the gravitons.
> 
> The mass engery of the gravitons amounts to
> 
> EmG = mG·c^2     (9)
> 
> The potential gravitational energy of the masses m1 und m2 amounts to
> 
> Epot = -G·Integral(from x to ∞) (m1·m2)/x^2 dx = −G·(m1·m2)/x     (10)
> 
> Equating the mass energy of the gravitons (eq. 9) with the potentially 
> gravitational energy (eq. 10) leads to
> 
> mG·c^2 = −G·(m1·m2)/x     (11)
> 
> Insertion into eq. 7 gives F·x = −G·(m1·m2)/x respectively
> 
> F = −G·(m1·m2)/x^2     (12)
> 
> The force, exerted by the graviton, corresponds exactly to the 
> gravitational force. Eq. 11 shows clearly, that the mass of the 
> gravitons is negative. This suggests to claim the next hypothesis.
> 
> 
> Hypothesis 3: The sum of all mass is zero.
> 
> The whole potential gravitational energy in the universe results from 
> the sum of the potential energy between every particle and every other 
> particle in the universe. This can be calculated by integration.
> 
> The universe shall be a spheric gas cloud with constant density ρu, 
> radius Ru and the mass Mu. A spherical shell with radius r and thickness 
> dr shall be examined. The potential energy between all particles on the 
> spherical shell ρu·dV among each other there is small of higher order 
> and hence can be neglected. For the calculation of the potential energy 
> between all particles on the spheric shell ρu·dV on the one part and all 
> other particles inside the sphere ρu·V on the other part, all particles 
> inside the sphere may summarized to one point.
> dEpot = −G·ρu^2·V·dV/r = −G·ρu^2·(1/r)·((4/3)·r^3·π)·(4·r^2·π·dr) = 
> −(16/3)·G·ρu^2·π^2·r^4·dr. Integration leads to Epot = 
> -(16/3)·G·ρu^2·π^2·Integral(from 0 to Ru) r^4·dr = 
> −(16/15)·G·ρu^2·π^2·Ru^5. The positive mass of the universe amounts to 
> Mu = Vu·ρu respectively
> 
> Mu = (4/3)·Ru^3·π·ρu     (13)
> 
> With this equation the potential energy can be written as follows:
> 
> Epot = −(3/5)·G·(Mu^2/Ru)     (14)
> 
> With EmG = Epot, with EmG from eq. 9 and Epot from eq. 14 the graviton
> mass can be calculated, while MG = Sum(mG) is representing the graviton 
> mass of the universe: MG·c^2 = −(3/5)·G·(Mu^2/Ru) respectively
> 
> MG = −(3/5)·G·Mu^2/(c^2·Ru)     (15)
> 
> The hypothesis
> 
> Sum(m) = MG + Mu = 0     (16)
> 
> with eq. 15 leads to the positive mass Mu of the universe: MG+Mu = 
> −(3/5)·G·Mu^2/(c^2·Ru)+Mu = 0 respectively
> 
> Mu = (5/3)·(c^2/G)·Ru = (5/3)·((299792458 m/s)^2/(6.67384E-11 
> m^3/(kg·s^2)))·4.26E26 m ≈ 1.0E54 kg     (17)
> 
> respectively to the radius of the universe dependent to its mass
> 
> Ru = (3/5)·Mu·(G/c^2)     (18)
> 
> Is the radius of the universe normalized by the Schwarzschild radius rS 
> = 2·Mu·(G/c^2), it amounts to
> 
> Ru = 0.3·rS     (19)
> 
> 
> Discussion:
> 
> The 1st hypothesis, that the gravitons are moving at the speed of light 
> between two masses back and forth, by the law of conservation of 
> momentum leads to the average force F ∼ 1/x^2 onto the two masses. The 
> 2nd hypothesis, that the potential energy of the gravitation is found in 
> the mass of the gravitons, leads to the result, that the gravitational 
> force onto both masses is caused exactly by
> the momentum exchange with the gravitons. The 3rd hypothesis, that the 
> sum of all masses, i. e. the sum of positive masses and negative 
> graviton masses in the universe is zero, leads to a plausible sum of the 
> positive mass Mu of the universe.
> 
> One problem is the mass of the gravitons, because it is negative und it 
> hasn't yet been observed any negative mass.
> 
> In order to be able to integrate eq. 5, it was assumed v<<c for 
> simplifiation. The transition to the in nitesimal consideration 
> represents no real restriction, when taking into account, that gravitons 
> from every particle in the universe are hitting mass m2 and hence a 
> continuous particle fl ow can be assumed. However the factor (c−v)/(c+v) 
> in eq. 5 would lead by exact calculation to the result, that the 
> gravitational force would be dependent to a former velocity v and hence 
> wouldn't be a potential  field no more. Further v<<c does not match to 
> the very fast expanding universe.
> It was assumed, that the universe is a continuum with constant density.
> Thereby an error was made, because the agglomeration of the galaxies and 
> the stars was neglected. I have estimated this error and it is very 
> small. Hence the positive mass of the universe has been calculated
> with constant density for better clarity.
> 
> 
> Used variables:
> 
> m1, m2 [kg]     positive mass
> Mu [kg]     sum of all positive mass in the universe
> mG [kg]     negative mass of a graviton
> MG [kg]     sum of all negative mass of all gravitons in the universe
> V [m^3]     volume
> ρu [kg/m^3]     density of the positive mass in the universe
> x [m]     distance between two positive masses
> r [m]     distance to the center of the universe
> v [m/s]     velocity of the mass m2
> ∆t [s]     fl ight time of the graviton back and forth
> P [kg·m/s]      momentum of the graviton
> F [N]     average force onto the mass m2
> Ru ≈ 4.26E26 m     radius of the universe
> Epot [J]     potential energy of the gravitation  field
> EmG [J]     mass energy of the graviton
> G = 6.67384E−11 m^3/(kg·s^2)     gravitational constant
> c = 299792458 m/s     speed of light
> rS = 2·Mu·(G/c^2)     Schwarzschild radius
> 
> 
> :-) Fritz Lasinger

Need a clear and consistent definition of your graviton particle properties before proceeding. Without this any theory can make bold statements without justification. Define your graviton first then stick too the limitation inherent in that definition.

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#382322

On Tuesday, April 26, 2016 at 5:12:07 PM UTC-7, John Heath wrote:
> On Saturday, April 23, 2016 at 5:13:33 PM UTC-4, Fritz Lasinger wrote:
> > Hello, I would like to present and discuss my new Momentum Theory for 
> > the Gravitation. I'm very interested in your opinions!
> > 
> > 
> > It will be shown, that gravitons only by momentum exchange exert a 
> > force, which is inversely proportional to their distance. If the energy, 
> > which is stored in the mass of the gravitons, corresponds to the 
> > potential energy of the attracting masses, the
> > gravitons cause exactly the gravitational force by the momentum 
> > exchange. Thereby the graviton mass becomes negative. Finally the 
> > assumption, that the sum of all positiv masses and negative graviton 
> > masses in the universe is zero, leads to a plausible positive mass of 
> > the universe of Mu ≈ 1.0·10^54 kg.
> > 
> > 
> > Introduction
> > 
> > In the following three hypotheses will be presented and it will be 
> > tried, to develop out of them the law of gravitation and the mass of the 
> > universe. The three hypotheses are:
> > 
> > 1. A so called graviton exists with the mass mG, which is moving at the
> > speed of light c between two masses m1 und m2 back and forth.
> > 
> > 2. The potential energy of the gravitation is found in the mass of the 
> > gravitons.
> > 
> > 3. The sum of all mass is zero.
> > 
> > 
> > Hypothesis 1: A so called graviton exists with the mass mG, which is 
> > moving at the speed of light c between two masses m1 and m2 back and forth.
> > 
> > An observer is moving with the same speed as mass m1. The mass m2 shall
> > have the distance x to the mass m1 and moves away from it with the speed 
> > v. The observer sees the graviton, moving toward the mass m2 with the 
> > relative momentum P = mG·(c−v). It fulfils a fully elastic impact and 
> > after that it moves backwards away from the mass m2 with the same 
> > relative momentum P = (mG+∆mG)·(c+v). Since the graviton cannot get 
> > slower, its mass is changing by ∆mG. The amount of the momentum has 
> > remained unchanged:
> > mG·(c−v) = (mG+∆mG)·(c+v) respectively ∆mG = mG·((c−v)/(c+v)-1)
> > 
> > ∆mG/mG = −2v/(c+v)     (1)
> > 
> > The graviton transfers the following momentum to the mass m2:
> > ∆P = mG·(c−v)+(mG+∆mG)·(c+v)
> > 
> > ∆P = 2·mG·c+∆mG·(c+v)     (2)
> > 
> > The fl ight time back and forth follows from ∆t/2·(c−v) = x
> > 
> > ∆t = 2x/(c−v)     (3)
> > 
> > Meanwhile the distance between m1 and m2 increases by
> > ∆x = v·∆t = 2x·v/(c−v).
> > 
> > ∆x/x = 2v/(c-v)     (4)
> > 
> > (1)/(4) gives the change of the graviton mass by the movement ∆x of mass m2:
> > (∆mG/mG)·(2v/(c-v)) = -(∆x/x)·2v/(c+v) respectively
> > 
> > ∆mG/mG = -((c-v)/(c+v))·(∆x/x)     (5)
> > 
> > To be able to integrate eq. 5, it is assumed v<<c for simpli cation. 
> > Thereby follows dmG/mg = -dx/x -> Integral(dmG/mG)=-Integral(dx/x) 
> > respectively
> > 
> > mG = mG0·x0/x     (6)
> > 
> > The average momentum fl ow, i. e. the average force, exerting on mass m2,
> > results from the transfered momentum (eq. 2), devided by the time (eq. 
> > 3) F = ∆P/∆t = ((2mG·c+∆mG·(c+v))/(2x))·(c−v). The equation is simpli ed 
> > by v<<c. Thereby follows F = mG·c²/x respectively
> > 
> > F·x = mG·c^2     (7)
> > 
> > Insertion of equation (6) leads to
> > 
> > F = mG0·c^2·x0/x^2     (8)
> > 
> > With F0 = mG0·c^2/x0 follows F = F0·x0^2/x^2. Hence the graviton exerts 
> > an average force F ∼ 1/x^2 to the masses m1 and m2, when it is assumed 
> > v<<c for simpli cation.
> > 
> > Eq. 7 suggests to claim the next hypothesis:
> > 
> > 
> > Hypothesis 2: The potential energy of the gravitation is
> > found in the mass of the gravitons.
> > 
> > The mass engery of the gravitons amounts to
> > 
> > EmG = mG·c^2     (9)
> > 
> > The potential gravitational energy of the masses m1 und m2 amounts to
> > 
> > Epot = -G·Integral(from x to ∞) (m1·m2)/x^2 dx = −G·(m1·m2)/x     (10)
> > 
> > Equating the mass energy of the gravitons (eq. 9) with the potentially 
> > gravitational energy (eq. 10) leads to
> > 
> > mG·c^2 = −G·(m1·m2)/x     (11)
> > 
> > Insertion into eq. 7 gives F·x = −G·(m1·m2)/x respectively
> > 
> > F = −G·(m1·m2)/x^2     (12)
> > 
> > The force, exerted by the graviton, corresponds exactly to the 
> > gravitational force. Eq. 11 shows clearly, that the mass of the 
> > gravitons is negative. This suggests to claim the next hypothesis.
> > 
> > 
> > Hypothesis 3: The sum of all mass is zero.
> > 
> > The whole potential gravitational energy in the universe results from 
> > the sum of the potential energy between every particle and every other 
> > particle in the universe. This can be calculated by integration.
> > 
> > The universe shall be a spheric gas cloud with constant density ρu, 
> > radius Ru and the mass Mu. A spherical shell with radius r and thickness 
> > dr shall be examined. The potential energy between all particles on the 
> > spherical shell ρu·dV among each other there is small of higher order 
> > and hence can be neglected. For the calculation of the potential energy 
> > between all particles on the spheric shell ρu·dV on the one part and all 
> > other particles inside the sphere ρu·V on the other part, all particles 
> > inside the sphere may summarized to one point.
> > dEpot = −G·ρu^2·V·dV/r = −G·ρu^2·(1/r)·((4/3)·r^3·π)·(4·r^2·π·dr) = 
> > −(16/3)·G·ρu^2·π^2·r^4·dr. Integration leads to Epot = 
> > -(16/3)·G·ρu^2·π^2·Integral(from 0 to Ru) r^4·dr = 
> > −(16/15)·G·ρu^2·π^2·Ru^5. The positive mass of the universe amounts to 
> > Mu = Vu·ρu respectively
> > 
> > Mu = (4/3)·Ru^3·π·ρu     (13)
> > 
> > With this equation the potential energy can be written as follows:
> > 
> > Epot = −(3/5)·G·(Mu^2/Ru)     (14)
> > 
> > With EmG = Epot, with EmG from eq. 9 and Epot from eq. 14 the graviton
> > mass can be calculated, while MG = Sum(mG) is representing the graviton 
> > mass of the universe: MG·c^2 = −(3/5)·G·(Mu^2/Ru) respectively
> > 
> > MG = −(3/5)·G·Mu^2/(c^2·Ru)     (15)
> > 
> > The hypothesis
> > 
> > Sum(m) = MG + Mu = 0     (16)
> > 
> > with eq. 15 leads to the positive mass Mu of the universe: MG+Mu = 
> > −(3/5)·G·Mu^2/(c^2·Ru)+Mu = 0 respectively
> > 
> > Mu = (5/3)·(c^2/G)·Ru = (5/3)·((299792458 m/s)^2/(6.67384E-11 
> > m^3/(kg·s^2)))·4.26E26 m ≈ 1.0E54 kg     (17)
> > 
> > respectively to the radius of the universe dependent to its mass
> > 
> > Ru = (3/5)·Mu·(G/c^2)     (18)
> > 
> > Is the radius of the universe normalized by the Schwarzschild radius rS 
> > = 2·Mu·(G/c^2), it amounts to
> > 
> > Ru = 0.3·rS     (19)
> > 
> > 
> > Discussion:
> > 
> > The 1st hypothesis, that the gravitons are moving at the speed of light 
> > between two masses back and forth, by the law of conservation of 
> > momentum leads to the average force F ∼ 1/x^2 onto the two masses. The 
> > 2nd hypothesis, that the potential energy of the gravitation is found in 
> > the mass of the gravitons, leads to the result, that the gravitational 
> > force onto both masses is caused exactly by
> > the momentum exchange with the gravitons. The 3rd hypothesis, that the 
> > sum of all masses, i. e. the sum of positive masses and negative 
> > graviton masses in the universe is zero, leads to a plausible sum of the 
> > positive mass Mu of the universe.
> > 
> > One problem is the mass of the gravitons, because it is negative und it 
> > hasn't yet been observed any negative mass.
> > 
> > In order to be able to integrate eq. 5, it was assumed v<<c for 
> > simplifiation. The transition to the in nitesimal consideration 
> > represents no real restriction, when taking into account, that gravitons 
> > from every particle in the universe are hitting mass m2 and hence a 
> > continuous particle fl ow can be assumed. However the factor (c−v)/(c+v) 
> > in eq. 5 would lead by exact calculation to the result, that the 
> > gravitational force would be dependent to a former velocity v and hence 
> > wouldn't be a potential  field no more. Further v<<c does not match to 
> > the very fast expanding universe.
> > It was assumed, that the universe is a continuum with constant density.
> > Thereby an error was made, because the agglomeration of the galaxies and 
> > the stars was neglected. I have estimated this error and it is very 
> > small. Hence the positive mass of the universe has been calculated
> > with constant density for better clarity.
> > 
> > 
> > Used variables:
> > 
> > m1, m2 [kg]     positive mass
> > Mu [kg]     sum of all positive mass in the universe
> > mG [kg]     negative mass of a graviton
> > MG [kg]     sum of all negative mass of all gravitons in the universe
> > V [m^3]     volume
> > ρu [kg/m^3]     density of the positive mass in the universe
> > x [m]     distance between two positive masses
> > r [m]     distance to the center of the universe
> > v [m/s]     velocity of the mass m2
> > ∆t [s]     fl ight time of the graviton back and forth
> > P [kg·m/s]      momentum of the graviton
> > F [N]     average force onto the mass m2
> > Ru ≈ 4.26E26 m     radius of the universe
> > Epot [J]     potential energy of the gravitation  field
> > EmG [J]     mass energy of the graviton
> > G = 6.67384E−11 m^3/(kg·s^2)     gravitational constant
> > c = 299792458 m/s     speed of light
> > rS = 2·Mu·(G/c^2)     Schwarzschild radius
> > 
> > 
> > :-) Fritz Lasinger
> 
> Need a clear and consistent definition of your graviton particle properties before proceeding. Without this any theory can make bold statements without justification. Define your graviton first then stick too the limitation inherent in that definition.

How about the particle is the real 
particle and gravity is fall gravity 
and unifies with the strong nuclear force.

The graviton here (force carrier of gravity) 
is the same as the particle.

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#382323

On Tuesday, April 26, 2016 at 9:15:08 PM UTC-7, Ross A. Finlayson wrote:
> On Tuesday, April 26, 2016 at 5:12:07 PM UTC-7, John Heath wrote:
> > On Saturday, April 23, 2016 at 5:13:33 PM UTC-4, Fritz Lasinger wrote:
> > > Hello, I would like to present and discuss my new Momentum Theory for 
> > > the Gravitation. I'm very interested in your opinions!
> > > 
> > > 
> > > It will be shown, that gravitons only by momentum exchange exert a 
> > > force, which is inversely proportional to their distance. If the energy, 
> > > which is stored in the mass of the gravitons, corresponds to the 
> > > potential energy of the attracting masses, the
> > > gravitons cause exactly the gravitational force by the momentum 
> > > exchange. Thereby the graviton mass becomes negative. Finally the 
> > > assumption, that the sum of all positiv masses and negative graviton 
> > > masses in the universe is zero, leads to a plausible positive mass of 
> > > the universe of Mu ≈ 1.0·10^54 kg.
> > > 
> > > 
> > > Introduction
> > > 
> > > In the following three hypotheses will be presented and it will be 
> > > tried, to develop out of them the law of gravitation and the mass of the 
> > > universe. The three hypotheses are:
> > > 
> > > 1. A so called graviton exists with the mass mG, which is moving at the
> > > speed of light c between two masses m1 und m2 back and forth.
> > > 
> > > 2. The potential energy of the gravitation is found in the mass of the 
> > > gravitons.
> > > 
> > > 3. The sum of all mass is zero.
> > > 
> > > 
> > > Hypothesis 1: A so called graviton exists with the mass mG, which is 
> > > moving at the speed of light c between two masses m1 and m2 back and forth.
> > > 
> > > An observer is moving with the same speed as mass m1. The mass m2 shall
> > > have the distance x to the mass m1 and moves away from it with the speed 
> > > v. The observer sees the graviton, moving toward the mass m2 with the 
> > > relative momentum P = mG·(c−v). It fulfils a fully elastic impact and 
> > > after that it moves backwards away from the mass m2 with the same 
> > > relative momentum P = (mG+∆mG)·(c+v). Since the graviton cannot get 
> > > slower, its mass is changing by ∆mG. The amount of the momentum has 
> > > remained unchanged:
> > > mG·(c−v) = (mG+∆mG)·(c+v) respectively ∆mG = mG·((c−v)/(c+v)-1)
> > > 
> > > ∆mG/mG = −2v/(c+v)     (1)
> > > 
> > > The graviton transfers the following momentum to the mass m2:
> > > ∆P = mG·(c−v)+(mG+∆mG)·(c+v)
> > > 
> > > ∆P = 2·mG·c+∆mG·(c+v)     (2)
> > > 
> > > The fl ight time back and forth follows from ∆t/2·(c−v) = x
> > > 
> > > ∆t = 2x/(c−v)     (3)
> > > 
> > > Meanwhile the distance between m1 and m2 increases by
> > > ∆x = v·∆t = 2x·v/(c−v).
> > > 
> > > ∆x/x = 2v/(c-v)     (4)
> > > 
> > > (1)/(4) gives the change of the graviton mass by the movement ∆x of mass m2:
> > > (∆mG/mG)·(2v/(c-v)) = -(∆x/x)·2v/(c+v) respectively
> > > 
> > > ∆mG/mG = -((c-v)/(c+v))·(∆x/x)     (5)
> > > 
> > > To be able to integrate eq. 5, it is assumed v<<c for simpli cation. 
> > > Thereby follows dmG/mg = -dx/x -> Integral(dmG/mG)=-Integral(dx/x) 
> > > respectively
> > > 
> > > mG = mG0·x0/x     (6)
> > > 
> > > The average momentum fl ow, i. e. the average force, exerting on mass m2,
> > > results from the transfered momentum (eq. 2), devided by the time (eq. 
> > > 3) F = ∆P/∆t = ((2mG·c+∆mG·(c+v))/(2x))·(c−v). The equation is simpli ed 
> > > by v<<c. Thereby follows F = mG·c²/x respectively
> > > 
> > > F·x = mG·c^2     (7)
> > > 
> > > Insertion of equation (6) leads to
> > > 
> > > F = mG0·c^2·x0/x^2     (8)
> > > 
> > > With F0 = mG0·c^2/x0 follows F = F0·x0^2/x^2. Hence the graviton exerts 
> > > an average force F ∼ 1/x^2 to the masses m1 and m2, when it is assumed 
> > > v<<c for simpli cation.
> > > 
> > > Eq. 7 suggests to claim the next hypothesis:
> > > 
> > > 
> > > Hypothesis 2: The potential energy of the gravitation is
> > > found in the mass of the gravitons.
> > > 
> > > The mass engery of the gravitons amounts to
> > > 
> > > EmG = mG·c^2     (9)
> > > 
> > > The potential gravitational energy of the masses m1 und m2 amounts to
> > > 
> > > Epot = -G·Integral(from x to ∞) (m1·m2)/x^2 dx = −G·(m1·m2)/x     (10)
> > > 
> > > Equating the mass energy of the gravitons (eq. 9) with the potentially 
> > > gravitational energy (eq. 10) leads to
> > > 
> > > mG·c^2 = −G·(m1·m2)/x     (11)
> > > 
> > > Insertion into eq. 7 gives F·x = −G·(m1·m2)/x respectively
> > > 
> > > F = −G·(m1·m2)/x^2     (12)
> > > 
> > > The force, exerted by the graviton, corresponds exactly to the 
> > > gravitational force. Eq. 11 shows clearly, that the mass of the 
> > > gravitons is negative. This suggests to claim the next hypothesis.
> > > 
> > > 
> > > Hypothesis 3: The sum of all mass is zero.
> > > 
> > > The whole potential gravitational energy in the universe results from 
> > > the sum of the potential energy between every particle and every other 
> > > particle in the universe. This can be calculated by integration.
> > > 
> > > The universe shall be a spheric gas cloud with constant density ρu, 
> > > radius Ru and the mass Mu. A spherical shell with radius r and thickness 
> > > dr shall be examined. The potential energy between all particles on the 
> > > spherical shell ρu·dV among each other there is small of higher order 
> > > and hence can be neglected. For the calculation of the potential energy 
> > > between all particles on the spheric shell ρu·dV on the one part and all 
> > > other particles inside the sphere ρu·V on the other part, all particles 
> > > inside the sphere may summarized to one point.
> > > dEpot = −G·ρu^2·V·dV/r = −G·ρu^2·(1/r)·((4/3)·r^3·π)·(4·r^2·π·dr) = 
> > > −(16/3)·G·ρu^2·π^2·r^4·dr. Integration leads to Epot = 
> > > -(16/3)·G·ρu^2·π^2·Integral(from 0 to Ru) r^4·dr = 
> > > −(16/15)·G·ρu^2·π^2·Ru^5. The positive mass of the universe amounts to 
> > > Mu = Vu·ρu respectively
> > > 
> > > Mu = (4/3)·Ru^3·π·ρu     (13)
> > > 
> > > With this equation the potential energy can be written as follows:
> > > 
> > > Epot = −(3/5)·G·(Mu^2/Ru)     (14)
> > > 
> > > With EmG = Epot, with EmG from eq. 9 and Epot from eq. 14 the graviton
> > > mass can be calculated, while MG = Sum(mG) is representing the graviton 
> > > mass of the universe: MG·c^2 = −(3/5)·G·(Mu^2/Ru) respectively
> > > 
> > > MG = −(3/5)·G·Mu^2/(c^2·Ru)     (15)
> > > 
> > > The hypothesis
> > > 
> > > Sum(m) = MG + Mu = 0     (16)
> > > 
> > > with eq. 15 leads to the positive mass Mu of the universe: MG+Mu = 
> > > −(3/5)·G·Mu^2/(c^2·Ru)+Mu = 0 respectively
> > > 
> > > Mu = (5/3)·(c^2/G)·Ru = (5/3)·((299792458 m/s)^2/(6.67384E-11 
> > > m^3/(kg·s^2)))·4.26E26 m ≈ 1.0E54 kg     (17)
> > > 
> > > respectively to the radius of the universe dependent to its mass
> > > 
> > > Ru = (3/5)·Mu·(G/c^2)     (18)
> > > 
> > > Is the radius of the universe normalized by the Schwarzschild radius rS 
> > > = 2·Mu·(G/c^2), it amounts to
> > > 
> > > Ru = 0.3·rS     (19)
> > > 
> > > 
> > > Discussion:
> > > 
> > > The 1st hypothesis, that the gravitons are moving at the speed of light 
> > > between two masses back and forth, by the law of conservation of 
> > > momentum leads to the average force F ∼ 1/x^2 onto the two masses. The 
> > > 2nd hypothesis, that the potential energy of the gravitation is found in 
> > > the mass of the gravitons, leads to the result, that the gravitational 
> > > force onto both masses is caused exactly by
> > > the momentum exchange with the gravitons. The 3rd hypothesis, that the 
> > > sum of all masses, i. e. the sum of positive masses and negative 
> > > graviton masses in the universe is zero, leads to a plausible sum of the 
> > > positive mass Mu of the universe.
> > > 
> > > One problem is the mass of the gravitons, because it is negative und it 
> > > hasn't yet been observed any negative mass.
> > > 
> > > In order to be able to integrate eq. 5, it was assumed v<<c for 
> > > simplifiation. The transition to the in nitesimal consideration 
> > > represents no real restriction, when taking into account, that gravitons 
> > > from every particle in the universe are hitting mass m2 and hence a 
> > > continuous particle fl ow can be assumed. However the factor (c−v)/(c+v) 
> > > in eq. 5 would lead by exact calculation to the result, that the 
> > > gravitational force would be dependent to a former velocity v and hence 
> > > wouldn't be a potential  field no more. Further v<<c does not match to 
> > > the very fast expanding universe.
> > > It was assumed, that the universe is a continuum with constant density.
> > > Thereby an error was made, because the agglomeration of the galaxies and 
> > > the stars was neglected. I have estimated this error and it is very 
> > > small. Hence the positive mass of the universe has been calculated
> > > with constant density for better clarity.
> > > 
> > > 
> > > Used variables:
> > > 
> > > m1, m2 [kg]     positive mass
> > > Mu [kg]     sum of all positive mass in the universe
> > > mG [kg]     negative mass of a graviton
> > > MG [kg]     sum of all negative mass of all gravitons in the universe
> > > V [m^3]     volume
> > > ρu [kg/m^3]     density of the positive mass in the universe
> > > x [m]     distance between two positive masses
> > > r [m]     distance to the center of the universe
> > > v [m/s]     velocity of the mass m2
> > > ∆t [s]     fl ight time of the graviton back and forth
> > > P [kg·m/s]      momentum of the graviton
> > > F [N]     average force onto the mass m2
> > > Ru ≈ 4.26E26 m     radius of the universe
> > > Epot [J]     potential energy of the gravitation  field
> > > EmG [J]     mass energy of the graviton
> > > G = 6.67384E−11 m^3/(kg·s^2)     gravitational constant
> > > c = 299792458 m/s     speed of light
> > > rS = 2·Mu·(G/c^2)     Schwarzschild radius
> > > 
> > > 
> > > :-) Fritz Lasinger
> > 
> > Need a clear and consistent definition of your graviton particle properties before proceeding. Without this any theory can make bold statements without justification. Define your graviton first then stick too the limitation inherent in that definition.
> 
> How about the particle is the real 
> particle and gravity is fall gravity 
> and unifies with the strong nuclear force.
> 
> The graviton here (force carrier of gravity) 
> is the same as the particle.

(This is with that the "speed of gravity" 
is instantaneous, not 'c', which is light 
speed and the absolute and relative limit 
on the apparent velocity of ordinary matter, 
and light, with an aether background, and 
gravity is a continuous field of fall gravity.)

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#382474

On 27.04.2016 02:12, John Heath wrote:
> On Saturday, April 23, 2016 at 5:13:33 PM UTC-4, Fritz Lasinger wrote:
>> Hello, I would like to present and discuss my new Momentum Theory for
>> the Gravitation. I'm very interested in your opinions!
>>
>>
>> It will be shown, that gravitons only by momentum exchange exert a
>> force, which is inversely proportional to their distance. If the energy,
>> which is stored in the mass of the gravitons, corresponds to the
>> potential energy of the attracting masses, the
>> gravitons cause exactly the gravitational force by the momentum
>> exchange. Thereby the graviton mass becomes negative. Finally the
>> assumption, that the sum of all positiv masses and negative graviton
>> masses in the universe is zero, leads to a plausible positive mass of
>> the universe of Mu ≈ 1.0·10^54 kg.
>>
>>
>> Introduction
>>
>> In the following three hypotheses will be presented and it will be
>> tried, to develop out of them the law of gravitation and the mass of the
>> universe. The three hypotheses are:
>>
>> 1. A so called graviton exists with the mass mG, which is moving at the
>> speed of light c between two masses m1 und m2 back and forth.
>>
>> 2. The potential energy of the gravitation is found in the mass of the
>> gravitons.
>>
>> 3. The sum of all mass is zero.
>>
>>
>> Hypothesis 1: A so called graviton exists with the mass mG, which is
>> moving at the speed of light c between two masses m1 and m2 back and forth.
>>
>> An observer is moving with the same speed as mass m1. The mass m2 shall
>> have the distance x to the mass m1 and moves away from it with the speed
>> v. The observer sees the graviton, moving toward the mass m2 with the
>> relative momentum P = mG·(c−v). It fulfils a fully elastic impact and
>> after that it moves backwards away from the mass m2 with the same
>> relative momentum P = (mG+∆mG)·(c+v). Since the graviton cannot get
>> slower, its mass is changing by ∆mG. The amount of the momentum has
>> remained unchanged:
>> mG·(c−v) = (mG+∆mG)·(c+v) respectively ∆mG = mG·((c−v)/(c+v)-1)
>>
>> ∆mG/mG = −2v/(c+v)     (1)
>>
>> The graviton transfers the following momentum to the mass m2:
>> ∆P = mG·(c−v)+(mG+∆mG)·(c+v)
>>
>> ∆P = 2·mG·c+∆mG·(c+v)     (2)
>>
>> The fl ight time back and forth follows from ∆t/2·(c−v) = x
>>
>> ∆t = 2x/(c−v)     (3)
>>
>> Meanwhile the distance between m1 and m2 increases by
>> ∆x = v·∆t = 2x·v/(c−v).
>>
>> ∆x/x = 2v/(c-v)     (4)
>>
>> (1)/(4) gives the change of the graviton mass by the movement ∆x of mass m2:
>> (∆mG/mG)·(2v/(c-v)) = -(∆x/x)·2v/(c+v) respectively
>>
>> ∆mG/mG = -((c-v)/(c+v))·(∆x/x)     (5)
>>
>> To be able to integrate eq. 5, it is assumed v<<c for simpli cation.
>> Thereby follows dmG/mg = -dx/x -> Integral(dmG/mG)=-Integral(dx/x)
>> respectively
>>
>> mG = mG0·x0/x     (6)
>>
>> The average momentum fl ow, i. e. the average force, exerting on mass m2,
>> results from the transfered momentum (eq. 2), devided by the time (eq.
>> 3) F = ∆P/∆t = ((2mG·c+∆mG·(c+v))/(2x))·(c−v). The equation is simpli ed
>> by v<<c. Thereby follows F = mG·c²/x respectively
>>
>> F·x = mG·c^2     (7)
>>
>> Insertion of equation (6) leads to
>>
>> F = mG0·c^2·x0/x^2     (8)
>>
>> With F0 = mG0·c^2/x0 follows F = F0·x0^2/x^2. Hence the graviton exerts
>> an average force F ∼ 1/x^2 to the masses m1 and m2, when it is assumed
>> v<<c for simpli cation.
>>
>> Eq. 7 suggests to claim the next hypothesis:
>>
>>
>> Hypothesis 2: The potential energy of the gravitation is
>> found in the mass of the gravitons.
>>
>> The mass engery of the gravitons amounts to
>>
>> EmG = mG·c^2     (9)
>>
>> The potential gravitational energy of the masses m1 und m2 amounts to
>>
>> Epot = -G·Integral(from x to ∞) (m1·m2)/x^2 dx = −G·(m1·m2)/x     (10)
>>
>> Equating the mass energy of the gravitons (eq. 9) with the potentially
>> gravitational energy (eq. 10) leads to
>>
>> mG·c^2 = −G·(m1·m2)/x     (11)
>>
>> Insertion into eq. 7 gives F·x = −G·(m1·m2)/x respectively
>>
>> F = −G·(m1·m2)/x^2     (12)
>>
>> The force, exerted by the graviton, corresponds exactly to the
>> gravitational force. Eq. 11 shows clearly, that the mass of the
>> gravitons is negative. This suggests to claim the next hypothesis.
>>
>>
>> Hypothesis 3: The sum of all mass is zero.
>>
>> The whole potential gravitational energy in the universe results from
>> the sum of the potential energy between every particle and every other
>> particle in the universe. This can be calculated by integration.
>>
>> The universe shall be a spheric gas cloud with constant density ρu,
>> radius Ru and the mass Mu. A spherical shell with radius r and thickness
>> dr shall be examined. The potential energy between all particles on the
>> spherical shell ρu·dV among each other there is small of higher order
>> and hence can be neglected. For the calculation of the potential energy
>> between all particles on the spheric shell ρu·dV on the one part and all
>> other particles inside the sphere ρu·V on the other part, all particles
>> inside the sphere may summarized to one point.
>> dEpot = −G·ρu^2·V·dV/r = −G·ρu^2·(1/r)·((4/3)·r^3·π)·(4·r^2·π·dr) =
>> −(16/3)·G·ρu^2·π^2·r^4·dr. Integration leads to Epot =
>> -(16/3)·G·ρu^2·π^2·Integral(from 0 to Ru) r^4·dr =
>> −(16/15)·G·ρu^2·π^2·Ru^5. The positive mass of the universe amounts to
>> Mu = Vu·ρu respectively
>>
>> Mu = (4/3)·Ru^3·π·ρu     (13)
>>
>> With this equation the potential energy can be written as follows:
>>
>> Epot = −(3/5)·G·(Mu^2/Ru)     (14)
>>
>> With EmG = Epot, with EmG from eq. 9 and Epot from eq. 14 the graviton
>> mass can be calculated, while MG = Sum(mG) is representing the graviton
>> mass of the universe: MG·c^2 = −(3/5)·G·(Mu^2/Ru) respectively
>>
>> MG = −(3/5)·G·Mu^2/(c^2·Ru)     (15)
>>
>> The hypothesis
>>
>> Sum(m) = MG + Mu = 0     (16)
>>
>> with eq. 15 leads to the positive mass Mu of the universe: MG+Mu =
>> −(3/5)·G·Mu^2/(c^2·Ru)+Mu = 0 respectively
>>
>> Mu = (5/3)·(c^2/G)·Ru = (5/3)·((299792458 m/s)^2/(6.67384E-11
>> m^3/(kg·s^2)))·4.26E26 m ≈ 1.0E54 kg     (17)
>>
>> respectively to the radius of the universe dependent to its mass
>>
>> Ru = (3/5)·Mu·(G/c^2)     (18)
>>
>> Is the radius of the universe normalized by the Schwarzschild radius rS
>> = 2·Mu·(G/c^2), it amounts to
>>
>> Ru = 0.3·rS     (19)
>>
>>
>> Discussion:
>>
>> The 1st hypothesis, that the gravitons are moving at the speed of light
>> between two masses back and forth, by the law of conservation of
>> momentum leads to the average force F ∼ 1/x^2 onto the two masses. The
>> 2nd hypothesis, that the potential energy of the gravitation is found in
>> the mass of the gravitons, leads to the result, that the gravitational
>> force onto both masses is caused exactly by
>> the momentum exchange with the gravitons. The 3rd hypothesis, that the
>> sum of all masses, i. e. the sum of positive masses and negative
>> graviton masses in the universe is zero, leads to a plausible sum of the
>> positive mass Mu of the universe.
>>
>> One problem is the mass of the gravitons, because it is negative und it
>> hasn't yet been observed any negative mass.
>>
>> In order to be able to integrate eq. 5, it was assumed v<<c for
>> simplifiation. The transition to the in nitesimal consideration
>> represents no real restriction, when taking into account, that gravitons
>> from every particle in the universe are hitting mass m2 and hence a
>> continuous particle fl ow can be assumed. However the factor (c−v)/(c+v)
>> in eq. 5 would lead by exact calculation to the result, that the
>> gravitational force would be dependent to a former velocity v and hence
>> wouldn't be a potential  field no more. Further v<<c does not match to
>> the very fast expanding universe.
>> It was assumed, that the universe is a continuum with constant density.
>> Thereby an error was made, because the agglomeration of the galaxies and
>> the stars was neglected. I have estimated this error and it is very
>> small. Hence the positive mass of the universe has been calculated
>> with constant density for better clarity.
>>
>>
>> Used variables:
>>
>> m1, m2 [kg]     positive mass
>> Mu [kg]     sum of all positive mass in the universe
>> mG [kg]     negative mass of a graviton
>> MG [kg]     sum of all negative mass of all gravitons in the universe
>> V [m^3]     volume
>> ρu [kg/m^3]     density of the positive mass in the universe
>> x [m]     distance between two positive masses
>> r [m]     distance to the center of the universe
>> v [m/s]     velocity of the mass m2
>> ∆t [s]     fl ight time of the graviton back and forth
>> P [kg·m/s]      momentum of the graviton
>> F [N]     average force onto the mass m2
>> Ru ≈ 4.26E26 m     radius of the universe
>> Epot [J]     potential energy of the gravitation  field
>> EmG [J]     mass energy of the graviton
>> G = 6.67384E−11 m^3/(kg·s^2)     gravitational constant
>> c = 299792458 m/s     speed of light
>> rS = 2·Mu·(G/c^2)     Schwarzschild radius
>>
>>
>> :-) Fritz Lasinger
>
> Need a clear and consistent definition of your graviton particle properties before proceeding. Without this any theory can make bold statements without justification. Define your graviton first then stick too the limitation inherent in that definition.
>
I'm only a mechanical engineer, so from my point of view the gravitons 
have the following properties:

- They have negative mass
- We probably cannot measure them, because the absolute value of the 
average mass is about 0.5E80 times smaller than the positive particle 
masses they belong to. (According to the number of particles in the 
universe, because there are about 10E80 particles with positiv mass and 
hence (10E80+1)*(10E80-2)/2≈0.5E160 gravitons with negativ mass and the 
particle ratio neg./pos. masses ist 0.5E160/10E80≈0.5E80 when the whole 
mass of the universe is about 0.)
- they move with speed of light
- they interact with positive masses with fully elastic impacts

I hope, this is enough for you :-)

Fritz

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#382509

Fritz Lasinger wrote:

> On 27.04.2016 02:12, John Heath wrote:
>> On Saturday, April 23, 2016 at 5:13:33 PM UTC-4, Fritz Lasinger wrote:
>>> An observer is moving with the same speed as mass m1. The mass m2 shall
>>> have the distance x to the mass m1 and moves away from it with the speed
>>> v. The observer sees the graviton, moving toward the mass m2 with the
>>> relative momentum P = mG·(c−v).

No, he does not.

> I'm only a mechanical engineer, so from my point of view the gravitons
> have the following properties:

(1) 
> - They have negative mass

(2)
> - We probably cannot measure them, because the absolute value of the
> average mass is about 0.5E80 times smaller than the positive particle
> masses they belong to.

A theory that is not falsifiable is not a theory at all.

> (According to the number of particles in the
> universe, because there are about 10E80 particles with positiv mass and
> hence (10E80+1)*(10E80-2)/2≈0.5E160 gravitons with negativ mass and the
> particle ratio neg./pos. masses ist 0.5E160/10E80≈0.5E80 when the whole
> mass of the universe is about 0.)

The total mass of the universe cannot be “about 0”; it would have ended in a 
Big Rip already (without ever producing the conditions necessary for the 
person making this up to come into existence) because according to the 
standard model of cosmology, confirmed by observation several times (at the 
latest, by the Planck Collaboration in 2015), ca. 71 % of the mass-energy of 
the universe comes from dark energy.

<https://en.wikipedia.org/wiki/Lambda-CDM_model>
<https://en.wikipedia.org/wiki/Dark_energy>

(3)
> - they move with speed of light

Inconsistent with proposition (1).  Particles with non-zero mass cannot move 
at the speed of light according to SR which you assume to be correct.  Put 
simply, an infinite amount of energy would be required for acceleration (see 
my other follow-up).

(4)
> - they interact with positive masses with fully elastic impacts
> 
> I hope, this is enough for you :-)

It is.


PointedEars
-- 
Heisenberg is out for a drive when he's stopped by a traffic cop. 
The officer asks him "Do you know how fast you were going?" 
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)

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#382536

> (1)
>> - They have negative mass
...
> (3)
>> - they move with speed of light
>
> Inconsistent with proposition (1).  Particles with non-zero mass cannot move
> at the speed of light according to SR which you assume to be correct.  Put
> simply, an infinite amount of energy would be required for acceleration (see
> my other follow-up).

Photons move with speed of light with non-zero mass, but they have no 
rest mass. This is, what I meant. Gravitons move with speed of light 
with non-zero mass, but without rest mass. If you mean, that they cannot 
decelerate, then they have to be absorbed by the positive mass, reducing 
whose energy and after a very short time they will be created again by 
increasing the energy of the positive mass again and are flying back. 
Just like a phonton which can be absorbed by an atom and after a short 
time emitted again, although the photon is moving with the speed of light.

:-) Fritz

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#382540

Fritz Lasinger wrote:

>> (1)
>>> - They have negative mass
> ...
>> (3)
>>> - they move with speed of light
>>
>> Inconsistent with proposition (1).  Particles with non-zero mass cannot
>> move at the speed of light according to SR which you assume to be
>> correct.  Put simply, an infinite amount of energy would be required for
>> acceleration (see my other follow-up).
> 
> Photons move with speed of light with non-zero mass

No, it is their *zero* mass that allows (and forces) photons to move at the 
speed of light in a medium in the first place (the Particle Data Group 
currently places the upper limit for its mass at less than 10⁻¹⁸ eV in 
natural units [1]).  Therefore, their total energy is E = p c = ℎ f, where p 
is the magnitude of their momentum, ℎ is the Planck constant, and f is the 
frequency of the corresponding electromagnetic wave.

[1] <http://pdg.lbl.gov/2015/tables/rpp2015-sum-gauge-higgs-bosons.pdf>

> but they have no rest mass.

In modern physics, there is only mass.  You are employing confusing, 
obsolete terminology introduced by Lewis and Tolman that even Einstein 
deprecated.  We have been over this before.  And ISTM you are confused
by it yourself.

<https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass>

> If you mean, that they cannot decelerate,

No, I do not mean that.  I said, in essence, that particles whose (rest) 
mass is not zero cannot be accelerated to reach the speed of light in the 
first place because in order to achieve that, an infinite amount of energy 
would be necessary.

> then they have to be absorbed by the positive mass,

Not even wrong.

> Just like a phonton which can be absorbed by an atom and after a short

A _photon_ is _not_ “absorbed by an atom”.  Instead, it can give off its 
energy to a subatomic particle, usually a shell electron (by which the 
photon ceases to exist), which is/are put into an excited state of higher 
energy, and might emit *another* photon after returning to its/their ground 
state of lower energy.  (Hence the reduced speed of light in optical thicker 
media.)

<https://en.wikipedia.org/wiki/Photoelectric_effect>

> time emitted again, although the photon is moving with the speed of light.

Irrelevant argument.

To summarize, I have showed by now that your "theory" is inconsistent with 
past and current physics, namely it is inconsistent with

- classical mechanics, in particular Newton’s law of universal gravitation;

- special relativity, in particular total energy, relativistic momentum,
  the velocity-addition formula, and the maximum speed of c;

- general relativity, in particular spacetime curvature at a distance from
  the source of the energy–momentum;

- quantum mechanics, in particular the uncertainty principle for energy;

- the standard model of cosmology, in particular dark energy;

- decades to centuries of experimental confirmation of all the theories 
  mentioned above, in their respective domains.

Also, your "theory" makes predictions that are not falsifiable, therefore it 
is not a scientific theory at all.

You should address these issues before you continue.  The other way lies 
madness – crackpottery.


PointedEars
-- 
Q: What did the nuclear physicist post on the laboratory door
   when he went camping?  
A: 'Gone fission'.
(from: WolframAlpha)

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#382543

On 30.04.2016 18:13, Thomas 'PointedEars' Lahn wrote:
> Fritz Lasinger wrote:
>
>>> (1)
>>>> - They have negative mass
>> ...
>>> (3)
>>>> - they move with speed of light
>>>
>>> Inconsistent with proposition (1).  Particles with non-zero mass cannot
>>> move at the speed of light according to SR which you assume to be
>>> correct.  Put simply, an infinite amount of energy would be required for
>>> acceleration (see my other follow-up).
>>
>> Photons move with speed of light with non-zero mass
>
> No, it is their *zero* mass that allows (and forces) photons to move at the
> speed of light in a medium in the first place (the Particle Data Group
> currently places the upper limit for its mass at less than 10⁻¹⁸ eV in
> natural units [1]).  Therefore, their total energy is E = p c = ℎ f, where p
> is the magnitude of their momentum, ℎ is the Planck constant, and f is the
> frequency of the corresponding electromagnetic wave.
>
> [1] <http://pdg.lbl.gov/2015/tables/rpp2015-sum-gauge-higgs-bosons.pdf>
>
>> but they have no rest mass.
>
> In modern physics, there is only mass.  You are employing confusing,
> obsolete terminology introduced by Lewis and Tolman that even Einstein
> deprecated.  We have been over this before.  And ISTM you are confused
> by it yourself.

photon:   v_P = c
graviton: v_G = c

photon:   m_P > 0
graviton: m_G < 0         (|m_G| << |m_P|)

photon:   E_P = p_P * c = +h * f_P > 0
graviton: E_G = p_G * c = -h * f_G < 0

photon:   m_P = p_P / c = +h * f_P / c^2 > 0
graviton: m_G = p_G / c = -h * f_G / c^2 < 0

Please tell me, why the graviton cannot move with the speed of light?

Fritz :-)

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#382564

Fritz Lasinger wrote:

> photon:   v_P = c
> graviton: v_G = c

So far, so good.

> photon:   m_P > 0

No, in the standard model, m_γ = 0. [As you can also see in the PDG summary 
I referred you to, “γ”, not “P”, is the standard symbol for the photon.]

> graviton: m_G < 0         (|m_G| << |m_P|)
> 
> photon:   E_P = p_P * c = +h * f_P > 0
> graviton: E_G = p_G * c = -h * f_G < 0
> 
> photon:   m_P = p_P / c = +h * f_P / c^2 > 0
> graviton: m_G = p_G / c = -h * f_G / c^2 < 0

Crackpottery, applying the Planck equation to a particle with non-zero mass, 
and calculating with either the negative of the Planck constant or a 
negative frequency.  Instead, for particles with non-zero mass, because of

  E = √((m c²)² + (p c)²)

the relation

  E(p = 0) = E₀ = ℎ f = ℎ c∕λ = m c²

holds only for their *rest* energy E₀ [λ then is the Compton wavelength of 
the particle].

> Please tell me, why the graviton cannot move with the speed of light?

It can and it does in the extended standard model; but for that one has to 
reject your proposition of its negative, i.e. non-zero mass.  AISB, only 
particles with zero mass can move at the speed of light, c.

As I indicated, in addition to inconsistencies with confirmed past and 
current physical theories, your so far failed attempt at proposing a 
physical theory is also *internally* inconsistent:

Either the graviton has zero mass (m = 0), then it *must* move at c, impart 
momentum p(f) = ℎ f∕c, carry a force of infinite range, and contribute 
nothing to the total *mass* of the universe (only to its total *energy*).  
[That is what we observe, most recently in the LIGO Scientific 
Collaboration’s results of GW150914.]

Or it has non-zero mass (m ≠ 0), then it *cannot* move at c, but only at 
v < c in *all* frames of reference; it would impart momentum
p(m, v) = m v∕√(1 – (v∕c)²), carries a force of finite range due to its 
limited lifetime ∆t ≈ ℏ∕(2 m c²) [that is \hbar – ℏ = ℎ∕2π; _not_ ℎ], and 
contributes something to the total mass of the universe.  [We do not observe 
that.]

Different from what you proposed, it *can not* be both.


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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#382537

...
>> Need a clear and consistent definition of your graviton particle
>> properties before proceeding. Without this any theory can make bold
>> statements without justification. Define your graviton first then
>> stick too the limitation inherent in that definition.
...
> 1 They have negative mass
> 2 We probably cannot measure them, because the absolute value of the
> average mass is about 0.5E80 times smaller than the positive particle
> masses they belong to. ...
> 3 they move with speed of light
> 4 they interact with positive masses with fully elastic impacts

more properties:

5. The ratio of wave length λ to the distance x is konstant for all 
Gravitons.
6. The wave length λ so so long, that we cannot observe any wave properties.
7. The lifetime of gravitons is very long, so we cannot observe 
disappearance or disappearance.

5: With λ=h/|p|=h/|mG*c| and mG=-G*m1*m2/(x*c^2) and m1=m2=mP (mP: 
proton mass - ) we get
λ=h/|-G*mp^2/(x*c^2)*c)| or
λ/x=(h*c)/(G*mP^2)=(6.626E-34Js*2.998E8ms^-1)/(6.67E-11m^3kg^-1s^-2*(1.67E-27kg))≈1.1E39=const.

6: λ/x≈1.1E39: The wave length of the gravitons is so much larger than 
the distans of the positive masses (for example protons), that one can 
say, we cannot observe wave properties.

7. The maximum flight distance before disappearing is r=c*∆t=h/(2*mG*c) 
according to PointedEars 24.04.2016, 10:57.
If follows the ratio of maximum flight distance r to the distance of the 
masses mP (assuming |mG| instead of mg):
r/x=(h*c)/(2*G*mP^2)=λ/(2x)≈0.5E39.
The flight distance is so much, that the gravitons can fly very very 
often between two masses.

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#382542

Fritz Lasinger wrote:

> 5. The ratio of wave length λ to the distance x is konstant for all
> Gravitons.
> 6. The wave length λ so so long, that we cannot observe any wave
> properties.

See below, respectively.

> 7. The lifetime of gravitons is very long, so we cannot
> observe disappearance or disappearance.

Inconsistent with its elaboration below.
 
> 5: With λ=h/|p|=h/|mG*c| and mG=-G*m1*m2/(x*c^2) and m1=m2=mP (mP:
> proton mass - ) we get
> λ=h/|-G*mp^2/(x*c^2)*c)| or
> 
λ/x=(h*c)/(G*mP^2)=(6.626E-34Js*2.998E8ms^-1)/(6.67E-11m^3kg^-1s^-2*(1.67E-27kg))≈1.1E39=const.

Not even wrong.
 
> 6: λ/x≈1.1E39: The wave length of the gravitons is so much larger than
> the distans of the positive masses (for example protons), that one can
> say, we cannot observe wave properties.

Inconsistent with modern quantum mechanics which states that *every* 
particle exhibits wave properties; it can be described by a characteristic 
wavefunction that is a solution of the Schrödinger equation.
 
> 7. The maximum flight distance before disappearing is r=c*∆t=h/(2*mG*c)
> according to PointedEars 24.04.2016, 10:57.

This is a gross misinterpretation of what I said there, and inconsistent 
with the summary above (if the particle can disappear after "flying" r(m), 
how can we not observe its disappearance?).  I said instead that the (*one-
way*) range of a force is limited by the (*rest*) mass of its virtual 
exchange particle.

> If follows the ratio of maximum flight distance r to the distance of the
> masses mP (assuming |mG| instead of mg):
> r/x=(h*c)/(2*G*mP^2)=λ/(2x)≈0.5E39.
> The flight distance is so much, that the gravitons can fly very very
> often between two masses.

Not even wrong.


Score adjusted

PointedEars
-- 
Q: What did the nuclear physicist post on the laboratory door
   when he went camping?  
A: 'Gone fission'.
(from: WolframAlpha

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