- Thank you received: 0
Requiem for Relativity
14 years 10 months ago #23921
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Oops[]To save you rummaging through your bookshelves, I used; which I think is right
a^(b + j c) = a^b(cos(c In a) + j sin(c In a))
Obviously square the results and take the sq root.
a^(b + j c) = a^b(cos(c In a) + j sin(c In a))
Obviously square the results and take the sq root.
Please Log in or Create an account to join the conversation.
14 years 10 months ago #23922
by Stoat
Replied by Stoat on topic Reply from Robert Turner
oh, while I'm on Joe, another question for you. I was looking at the idea of an f.m particle, which would have a wave, the cosine of the natural log of the lorentzian. This does give anti parallel wave movement but i couldn't see what any information could be transmitted by it.
Then I was looking at the "fractal" y =|x| could I just do the same with that cosine wave and say that the gaps in the half wave can sometimes hide a devil's step in them?
Then I was looking at the "fractal" y =|x| could I just do the same with that cosine wave and say that the gaps in the half wave can sometimes hide a devil's step in them?
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
14 years 10 months ago #23158
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, ...h = c^2 / b^2 is the best bet for the speed of gravity; b being the speed of gravity here. ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Comment by JK: I like it, that h is dimensionless, which is correct. I also like it that you are looking to mathematical number theory, to find dimensionless numbers. The Riemann zeta function, and prime numbers generally, are central to number theory, so I like that strategy of yours too: looking to prime numbers and the zeta function. I think there's something important here, but I don't have enough time to work on this idea as much as I want to.
I'm inclined to avoid complex numbers here, if possible. Consider that the fine structure constant seems to be 1/137.036, and 137 is prime. Also consider that the (dimensionless) electric-to-gravity ratio,
"X" = q^2 / (G*m^2)
where q & m are the electron charge & mass, resp., and G is the gravitational constant, is about
"X" = 4.1655 * 10^42
I find that (the base of the logarithm doesn't matter, because this is a ratio of logs) log(X)/log(137.036) = 19 + 1 - 1/19 - 1/19^2 -..., with only 0.004% error.
<br />Hi Joe, ...h = c^2 / b^2 is the best bet for the speed of gravity; b being the speed of gravity here. ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Comment by JK: I like it, that h is dimensionless, which is correct. I also like it that you are looking to mathematical number theory, to find dimensionless numbers. The Riemann zeta function, and prime numbers generally, are central to number theory, so I like that strategy of yours too: looking to prime numbers and the zeta function. I think there's something important here, but I don't have enough time to work on this idea as much as I want to.
I'm inclined to avoid complex numbers here, if possible. Consider that the fine structure constant seems to be 1/137.036, and 137 is prime. Also consider that the (dimensionless) electric-to-gravity ratio,
"X" = q^2 / (G*m^2)
where q & m are the electron charge & mass, resp., and G is the gravitational constant, is about
"X" = 4.1655 * 10^42
I find that (the base of the logarithm doesn't matter, because this is a ratio of logs) log(X)/log(137.036) = 19 + 1 - 1/19 - 1/19^2 -..., with only 0.004% error.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
14 years 10 months ago #23190
by Joe Keller
Replied by Joe Keller on topic Reply from
The Asteroid Resonance - Part 7
Theoretical basis of 5.13h rotation period. The "Planck time" is sqrt(hbar*G/c^5). Essentially, the 5.13h rotation, is to the Planck time, as gravity is to electricity.
The usual definition of the electric to gravity ratio, for the electron, is q^2/(G*m^2). I consider a "modified electric to gravity ratio" which counts only the gravitational attraction of the electric field to itself. For a maximally compressed electron, according to the Schroedinger wave theory explanation of the Heisenberg uncertainty principle, the mass-energy of the field is alpha times the mass-energy of the electron, where alpha is the fine structure constant. So, I multiply the electric to gravity ratio by 1/alpha^2.
A refinement, is to realize that when, through compression of the electron and its field, the total electron mass increases from m to m + m*alpha, the electron can be smaller by a factor (1+alpha), because more mass-energy is available to make shorter deBroglie waves. The field now isn't just m*alpha, it's (m + m*alpha) * alpha = m*(alpha + alpha^2). Successive such approximations lead to m*(alpha+alpha^2+alpha^3+...) = m*alpha/(1-alpha).
So, my "modified electric to gravity ratio" is q^2/(G*mprime^2), where mprime = m*alpha/(1-alpha). The "gravitational Planck time", is defined as "Planck time" * "modified electric to gravity ratio", and satisfies the equation:
"gravitational Planck time" * 2*pi / sqrt(2) = 5.1287hr.
The four asteroids showing alignment with Barbarossa at the end of the Mayan Long Count, have rotation periods from 5.1294 to 5.1655hr.
Alignment of a "quasi-Centaur" with Barbarossa. Duffard's 2009 arXiv.org paper (see earlier post) has a table at the end (after the references) of TNO rotation periods. The only TNO whose rotation period, as given there, might be 5.13h, is 2002 PN34 (asteroid #73480), whose rotation period is given as
"4.23 or 5.11) +/- 0.03hr".
This would seem to include the possibility of 5.11+0.02=5.13hr.
At 2012, this TNO (diam. ~ 100km) is indeed only about an hour of ecliptic longitude past opposition to Barbarossa. Online sources give various estimates of its period or semimajor axis, corresponding to periods from 170.5yr to 174.8yr.
The orbit is far from the ecliptic when it crosses Uranus or Neptune, but is near the ecliptic at perihelion < 4AU outside Saturn's orbit, so Saturn is the main perturbative influence. Resonance (6::1) with Saturn would give orbital period 176.75yr. The remainders on division into my "first", "second", and "third" periods (see previous posts) are 0.867, 0.994, and 0.838.
Theoretical basis of 5.13h rotation period. The "Planck time" is sqrt(hbar*G/c^5). Essentially, the 5.13h rotation, is to the Planck time, as gravity is to electricity.
The usual definition of the electric to gravity ratio, for the electron, is q^2/(G*m^2). I consider a "modified electric to gravity ratio" which counts only the gravitational attraction of the electric field to itself. For a maximally compressed electron, according to the Schroedinger wave theory explanation of the Heisenberg uncertainty principle, the mass-energy of the field is alpha times the mass-energy of the electron, where alpha is the fine structure constant. So, I multiply the electric to gravity ratio by 1/alpha^2.
A refinement, is to realize that when, through compression of the electron and its field, the total electron mass increases from m to m + m*alpha, the electron can be smaller by a factor (1+alpha), because more mass-energy is available to make shorter deBroglie waves. The field now isn't just m*alpha, it's (m + m*alpha) * alpha = m*(alpha + alpha^2). Successive such approximations lead to m*(alpha+alpha^2+alpha^3+...) = m*alpha/(1-alpha).
So, my "modified electric to gravity ratio" is q^2/(G*mprime^2), where mprime = m*alpha/(1-alpha). The "gravitational Planck time", is defined as "Planck time" * "modified electric to gravity ratio", and satisfies the equation:
"gravitational Planck time" * 2*pi / sqrt(2) = 5.1287hr.
The four asteroids showing alignment with Barbarossa at the end of the Mayan Long Count, have rotation periods from 5.1294 to 5.1655hr.
Alignment of a "quasi-Centaur" with Barbarossa. Duffard's 2009 arXiv.org paper (see earlier post) has a table at the end (after the references) of TNO rotation periods. The only TNO whose rotation period, as given there, might be 5.13h, is 2002 PN34 (asteroid #73480), whose rotation period is given as
"4.23 or 5.11) +/- 0.03hr".
This would seem to include the possibility of 5.11+0.02=5.13hr.
At 2012, this TNO (diam. ~ 100km) is indeed only about an hour of ecliptic longitude past opposition to Barbarossa. Online sources give various estimates of its period or semimajor axis, corresponding to periods from 170.5yr to 174.8yr.
The orbit is far from the ecliptic when it crosses Uranus or Neptune, but is near the ecliptic at perihelion < 4AU outside Saturn's orbit, so Saturn is the main perturbative influence. Resonance (6::1) with Saturn would give orbital period 176.75yr. The remainders on division into my "first", "second", and "third" periods (see previous posts) are 0.867, 0.994, and 0.838.
Please Log in or Create an account to join the conversation.
14 years 10 months ago #23159
by Stoat
Replied by Stoat on topic Reply from Robert Turner
Hi Joe, food for thought there. I've been looking at the esu as well. First thought being what's the force if we wanted two electrons touching each other, at twice the Compton wavelength of the electron. Now to get my speed of gravity here, I had to decrease the Compton wavelength to about 2.39E-12 so I stuck in a cos theta, as a temporary little patch. I think it's fair enough, as these two electrons are spinning.
Divide that force by the mass of an electron to get a deceleration of twice the speed of gravity. Integrate twice to get d = ut + bt^2 and take u as being the Fermi velocity for copper. The displacement d again being the Compton wavelength. So a positive time root and a negative time root on just either side of the charge number value. Personally I don't care much for the idea of negative time but as a heuristic device it's okay here.
Something else to note , it's a square law for gravity but the deceleration is pretty horrifying, it's down to sub light in less than a millimeter.
Now I chose the Fermi velocity of copper, for no other reason than it's a common metal. All of the Fermi velocities I've seen, are at about a hundredth of the speed of light. The fine structure constant springs to mind there as well.
Any way the upshot is, for the electron, esu force of about 1.01E-5 newtons,
f_esu /2b*c^2 = f_grav
where b is the speed of gravity.
For the emu we take the esu force, where once again, the radius is twice the Compton wavelength of the electron squared.
f_esu/c^2 = f_emu
Divide that force by the mass of an electron to get a deceleration of twice the speed of gravity. Integrate twice to get d = ut + bt^2 and take u as being the Fermi velocity for copper. The displacement d again being the Compton wavelength. So a positive time root and a negative time root on just either side of the charge number value. Personally I don't care much for the idea of negative time but as a heuristic device it's okay here.
Something else to note , it's a square law for gravity but the deceleration is pretty horrifying, it's down to sub light in less than a millimeter.
Now I chose the Fermi velocity of copper, for no other reason than it's a common metal. All of the Fermi velocities I've seen, are at about a hundredth of the speed of light. The fine structure constant springs to mind there as well.
Any way the upshot is, for the electron, esu force of about 1.01E-5 newtons,
f_esu /2b*c^2 = f_grav
where b is the speed of gravity.
For the emu we take the esu force, where once again, the radius is twice the Compton wavelength of the electron squared.
f_esu/c^2 = f_emu
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
14 years 10 months ago #23161
by Joe Keller
Replied by Joe Keller on topic Reply from
The Asteroid Resonance - Part 8
The Cassini space probe found that Saturn's innermost ring, D68, has a complicated shape, but usually lies within a few km of 67645km from Saturn's center (Hedman, Burns, Showalter et al, Icarus 2007). (D68's radius has increased slightly since the 67580 +/- 10km found by Voyager; see Showalter, 1996.) Comparison with Titan's orbit, using Kepler's law relating semimajor axis to period, implies an orbital period of 4.895h for ring D68, if its orbit is circular at the distance found by Cassini. (This neglects the gravitational effect of Saturn's bulge, and of ring and moon material). The same article by Hedman et al (sec. 7, p. 18) says that the shortest orbital period of ring material, is 5.25h. So, the accurate minimum period might be 5.13h. (Hedman et al remark that Saturn's true mean rotation period, ~10.5h, also is slightly uncertain, so there might be exact 2::1 resonance.)
Few eclipsing binary stars have orbital periods shorter than 5.13h. The closest to 5.13h, is AV Telescopii, with 5.16h. In Dec. 2012AD, this star will be at a heliocentric angle of 95.35deg from Barbarossa. Sky surveys show that within a few arcminutes of this star, many other stars lie along short arcs (whatever the significance, if any, of that might be). AV Telescopii is Visual mag 14.1-15.1, variable type E/KW a.k.a. E+KW, where "E" signifies "eclipsing binary" and "KW" signifies "yellow main-sequence primary with hotter subdwarf secondary" ( www.assa.org.au ); the "K" signifies "contact" (specifically, the first phase of mass exchange) not color spectral type K. If the primary is spectral type mid-G V with magnitude alone +15.1, then "spectral parallax" (i.e. distance estimate assuming average absolute magnitude for spectral type) gives 1000pc distance (or nearer if there is intervening dust). The star's mean proper motion corroborates this, suggesting ~400pc distance according to our Sun's apex motion. Though this star might be too distant to influence our solar system, it might show effects related to its 5.13h (or rather, 5.16h) resonance.
The Cassini space probe found that Saturn's innermost ring, D68, has a complicated shape, but usually lies within a few km of 67645km from Saturn's center (Hedman, Burns, Showalter et al, Icarus 2007). (D68's radius has increased slightly since the 67580 +/- 10km found by Voyager; see Showalter, 1996.) Comparison with Titan's orbit, using Kepler's law relating semimajor axis to period, implies an orbital period of 4.895h for ring D68, if its orbit is circular at the distance found by Cassini. (This neglects the gravitational effect of Saturn's bulge, and of ring and moon material). The same article by Hedman et al (sec. 7, p. 18) says that the shortest orbital period of ring material, is 5.25h. So, the accurate minimum period might be 5.13h. (Hedman et al remark that Saturn's true mean rotation period, ~10.5h, also is slightly uncertain, so there might be exact 2::1 resonance.)
Few eclipsing binary stars have orbital periods shorter than 5.13h. The closest to 5.13h, is AV Telescopii, with 5.16h. In Dec. 2012AD, this star will be at a heliocentric angle of 95.35deg from Barbarossa. Sky surveys show that within a few arcminutes of this star, many other stars lie along short arcs (whatever the significance, if any, of that might be). AV Telescopii is Visual mag 14.1-15.1, variable type E/KW a.k.a. E+KW, where "E" signifies "eclipsing binary" and "KW" signifies "yellow main-sequence primary with hotter subdwarf secondary" ( www.assa.org.au ); the "K" signifies "contact" (specifically, the first phase of mass exchange) not color spectral type K. If the primary is spectral type mid-G V with magnitude alone +15.1, then "spectral parallax" (i.e. distance estimate assuming average absolute magnitude for spectral type) gives 1000pc distance (or nearer if there is intervening dust). The star's mean proper motion corroborates this, suggesting ~400pc distance according to our Sun's apex motion. Though this star might be too distant to influence our solar system, it might show effects related to its 5.13h (or rather, 5.16h) resonance.
Please Log in or Create an account to join the conversation.
Time to create page: 0.530 seconds