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Requiem for Relativity
14 years 6 months ago #23895
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Em, I just noticed that barh is wrong in that paper, must have been a printer's typo. It should be 1.057E-27 erg sec
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14 years 6 months ago #23896
by Stoat
Replied by Stoat on topic Reply from Robert Turner
I've just realized that I've got t wrong as well. I was taking t as being the age of the universe in atomic units. From the Dirac cosmological model we can work out t from the equation, G = 1 / 6 pi rho t^2
Taking G as 6.7E-8 erg cm g^-2
rho = 4E-31 g cm^-2 (the mean density of the universe)
Then t = 1.4E 18 seconds. That makes the Dirac universe about three time bigger than current estimates.
What I think we should explore is the idea that both G and barh vary with time. We can have an expanding universe in G decreases over time, or we can have an expanding universe where barh increases. A steady state universe would have both happening.
Taking G as 6.7E-8 erg cm g^-2
rho = 4E-31 g cm^-2 (the mean density of the universe)
Then t = 1.4E 18 seconds. That makes the Dirac universe about three time bigger than current estimates.
What I think we should explore is the idea that both G and barh vary with time. We can have an expanding universe in G decreases over time, or we can have an expanding universe where barh increases. A steady state universe would have both happening.
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- Joe Keller
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14 years 6 months ago #23897
by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Stoat</i>
<br />Hi Joe, ... Now that m_o is taken to be the mass of the pion by Dirac and co. That's about 273 times the mass of an aelctron. So the ratio of m_o / 137 m_e is about 1.99 ...
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Nice work! This is a clue that the mass of the pion depends on some simple formula involving the fine structure constant.
In 2004, I modeled the heavy leptons, and mesons and baryons, as "electric black holes" using Weyl's idea, that the electric field could be treated as a very strong gravitational field, and general relativity applied. The muon is the simplest of these; the tauon was like a heavy muon with a certain special property. The mesons and baryons had to be modeled as two- or three-layer onions of charge (each layer a quark). In each case the "singularity surface" was at a quark (i.e. layer) boundary.
For the pion, for example, the mass assumes the smallest possible wave-mechanical spherical gaussian distribution and the innermost half of the mass is assigned 1/3 or 2/3 charge, the outermost, likewise 1/3 or 2/3 charge, always with a fixed charge/mass ratio in the substance of each quark, either of the same sign or opposite sign, so the charge adds to 0 or +/-1.
<br />Hi Joe, ... Now that m_o is taken to be the mass of the pion by Dirac and co. That's about 273 times the mass of an aelctron. So the ratio of m_o / 137 m_e is about 1.99 ...
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Nice work! This is a clue that the mass of the pion depends on some simple formula involving the fine structure constant.
In 2004, I modeled the heavy leptons, and mesons and baryons, as "electric black holes" using Weyl's idea, that the electric field could be treated as a very strong gravitational field, and general relativity applied. The muon is the simplest of these; the tauon was like a heavy muon with a certain special property. The mesons and baryons had to be modeled as two- or three-layer onions of charge (each layer a quark). In each case the "singularity surface" was at a quark (i.e. layer) boundary.
For the pion, for example, the mass assumes the smallest possible wave-mechanical spherical gaussian distribution and the innermost half of the mass is assigned 1/3 or 2/3 charge, the outermost, likewise 1/3 or 2/3 charge, always with a fixed charge/mass ratio in the substance of each quark, either of the same sign or opposite sign, so the charge adds to 0 or +/-1.
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14 years 6 months ago #23899
by Stoat
Replied by Stoat on topic Reply from Robert Turner
I wonder if we are sometimes using the number two in our equations when we should be using that mass of a pion divided by 137 times the mass of an electron?
Mass of pion 9.109534E-31 kg times 263.13 = 2.39699E-28 Divide that by 137 times the mass of electron to get 1.9206 Divide that into the number two, 1.0413
Now if we say that barh = c^2 / b^2 where b is the speed of gravity then we can also look at
barh = v^2 /c^2 They are vastly different in scale but proportionate. Then the velocity we get, is going to be something like 1E-9 something very very cold.
We can also pop the equation r = 2Gm /c^2 into the lorentzian equation, where r is equal to barh. The mass we get here is 7.09673E-8 So what's that in terms of the Planck mass? Let's try dividing it by pi. That's 2.2589E-8 Now that's a little high, so divide that by the Planck mass, which is 2.17644E-8 and that gives us 1.03792 which compares fairly nicely with that value of 1.0413 that we got earlier.
Mass of pion 9.109534E-31 kg times 263.13 = 2.39699E-28 Divide that by 137 times the mass of electron to get 1.9206 Divide that into the number two, 1.0413
Now if we say that barh = c^2 / b^2 where b is the speed of gravity then we can also look at
barh = v^2 /c^2 They are vastly different in scale but proportionate. Then the velocity we get, is going to be something like 1E-9 something very very cold.
We can also pop the equation r = 2Gm /c^2 into the lorentzian equation, where r is equal to barh. The mass we get here is 7.09673E-8 So what's that in terms of the Planck mass? Let's try dividing it by pi. That's 2.2589E-8 Now that's a little high, so divide that by the Planck mass, which is 2.17644E-8 and that gives us 1.03792 which compares fairly nicely with that value of 1.0413 that we got earlier.
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14 years 6 months ago #23901
by johnboy
Replied by johnboy on topic Reply from
This question is to Joe Keller: Sorry it is three years late. But the topic is still open, so I guess it's OK. On your post of 12 Feb 2007 you said:
"The CMB dipole is caused by the sun's retrograde, small cool brown dwarf, 0.019 solar mass, mag +18 companion at 360 AU distance in the positive dipole direction."
This was in the context of several posts dealing with the effect of various known (Neptune) and hypothetical objects on the temperature anisotropies in the CMB. You went on to discuss how the orbits of the objects should cause a shift in the direction of the various multipole moments of the CMB.
Now you also mentioned, unless I misread your intent, that the source of the CMB was a shell at about 53 AU, where phenomenon of a hitherto unsuspected nature was at work.
If this is the case,attempting to demonstrate that the CMB is sourced at this distence by fitting the data on movement of the direction of the multipoles to the gravitational effect of a hypothetical object, while feasible, is more complicated then it needs to be.
If the CMB is coming from a source at ~50 AU it seems to me that the dipole directions would show a MAJOR parallax simply by using the Earths orbital baseline of 2 AU. Wouldn't the parallex be 2x arctan(1/52)=2.2 deg?
This seems to me to be a very much simpler and unambiguous way to determine if the CMB is coming from such a close source.
Apologies in advance if you already addressed this issue in another post. I only have read the first 4 pages of the 47 total.
"The CMB dipole is caused by the sun's retrograde, small cool brown dwarf, 0.019 solar mass, mag +18 companion at 360 AU distance in the positive dipole direction."
This was in the context of several posts dealing with the effect of various known (Neptune) and hypothetical objects on the temperature anisotropies in the CMB. You went on to discuss how the orbits of the objects should cause a shift in the direction of the various multipole moments of the CMB.
Now you also mentioned, unless I misread your intent, that the source of the CMB was a shell at about 53 AU, where phenomenon of a hitherto unsuspected nature was at work.
If this is the case,attempting to demonstrate that the CMB is sourced at this distence by fitting the data on movement of the direction of the multipoles to the gravitational effect of a hypothetical object, while feasible, is more complicated then it needs to be.
If the CMB is coming from a source at ~50 AU it seems to me that the dipole directions would show a MAJOR parallax simply by using the Earths orbital baseline of 2 AU. Wouldn't the parallex be 2x arctan(1/52)=2.2 deg?
This seems to me to be a very much simpler and unambiguous way to determine if the CMB is coming from such a close source.
Apologies in advance if you already addressed this issue in another post. I only have read the first 4 pages of the 47 total.
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14 years 6 months ago #24040
by Maurol
Replied by Maurol on topic Reply from Mauro Lacy
Hi all,
Some of you might be interested in the following paper:
<b>Dark Fluid: Towards a unification of empirical theories of galaxy rotation, Inflation and Dark Energy</b>
[url] arxiv.org/abs/0804.1588 [/url]
The math is extremely hard to understand, at least to me, but the general idea is similar to something I've expressed here in the past: gravity could be a force without a material cause, while accumulation of matter is just its consequence.
Best regards,
Mauro
Some of you might be interested in the following paper:
<b>Dark Fluid: Towards a unification of empirical theories of galaxy rotation, Inflation and Dark Energy</b>
[url] arxiv.org/abs/0804.1588 [/url]
The math is extremely hard to understand, at least to me, but the general idea is similar to something I've expressed here in the past: gravity could be a force without a material cause, while accumulation of matter is just its consequence.
Best regards,
Mauro
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