Requiem for Relativity

It has to be an approach from Joe.

Preview with mathematical details

Empirically, sqrt(kT/m) = H0 * c * P, to about 5% accuracy, maybe better. The left side of the equation is the isothermal speed of sound in an ideal gas, of any density, of protons (or hydrogen atoms or neutrons) at the CMB temperature. On the right side, H0 is the Hubble parameter; c is the speed of light; P is the Barbarossa period, ~ 6340 yr.

H0 * c equals, to about 10% accuracy, the anomalous Pioneer probe sunward acceleration (also observed for the Galileo and Ulysses probes). This suggests that an interplanetary proton (dark matter?) or atomic hydrogen gas, effectively at the CMB temperature (not solar radiation equilibrium temperature) is oscillating under the influence of the Pioneer anomalous sunward acceleration, with the Barbarossa period.

I'm solving the partial differential equations for the simplest such oscillation, a radial oscillation. Conservation of mass and momentum, as in the usual derivation of the wave equation for sound, give a system of two quasilinear first-order partial differential equations:

du/dt = -d(uv)/dr - 2*u*v/r

dv/dt = -v*dv/dr - a - K/u * du/dr

where "d" should be the curly "d" of partial differentiation; u = density; v = radial velocity; a is the Pioneer anomalous acceleration; K = k*T/m = mean square thermal velocity component along one axis, for T = the CMB temperature.

I could solve this system numerically using the Crank-Nicolson scheme (a standard second-order, unconditionally stable, finite difference scheme) with Gauss-Seidel iteration (see, inter alia, Strikwerda, Finite Difference Schemes and Partial Differential Equations). On the other hand, it is well known (see, inter alia, Courant & Hilbert, Methods of Mathematical Physics, vol. 2) that for hyperbolic systems like this, the "shock" fronts occur along "characteristic curves". Not all characteristic curves are shock fronts, but all shock fronts are characteristic curves. So, I would rather use a numerical method that gives the characteristic curve grid, instead of using a rectilinear grid. Then I will know radius as a function of time, along many characteristic curves, i.e. possible shock fronts.

For the system above, DJ Panov, Formulas for the Numerical Solution of Partial Differential Equations by the Method of Differences (Ungar, 1963; translation of 1951 USSR edition), ch. VII, pp. 103-107 gives the needed formulas to find the characteristic grid, numerically. I found two misprints in Panov's formulas. In eqn. (42) under "General Remarks" at the top of p. 103, the denominator in the second factor of the second term of the second line, should be dx, not dy. At the top of p. 104, the determinant for "A" should have a21 in the lower left corner, not b21.

However, textbooks oftener rewrite the above system as a second-order PDE in one unknown (this can be done using the equation d2(log(u))/dtdr = d2(log(u))/drdt ) before finding the characteristic grid. The needed formulas for this second-order case are in Street, Analysis and Solution of Partial Differential Equations, secs. 9.3.1 & 9.3.3; or, slightly more concisely and precisely, in G. Evans et al, Numerical Methods for Partial Differential Equations (Springer, 2000), sec. 3.3, pp. 80-83; I found a misprint in Evans' formulas: at the bottom of p. 82, in eqn. 3.3.18, the subscript of the first "g" should be "Q", not "P". G. Evans' book gives the same notation, same equations, and practically the same text, as GD Smith, Numerical Solution of Partial Differential Equations (Oxford 1965), beginning of ch. 4, pp. 98-102. GD Smith's textbook has several later editions and remains popular; the 2nd (1978) edition gives this material later in ch. 4, on pp. 161-166. I offer this much help, because I hope that experts in applied mathematics will address this problem themselves.

Elimination of u gives a second order quasi-linear hyperbolic PDE for radial velocity:

d2v/dt2 - (K-v^2)*d2v/dr2 + 2*v*d2v/dtdr

+ 2*dv/dr*dv/dt + 2*v*(dv/dr)^2 - (2*K/r - a) * dv/dr + 2*K*v/r^2 = 0

This equation is hyperbolic because the discriminant = 4*K > 0 independent of v. Scaling this equation, that is, changing units of time and distance so that "K" and "a" become unity, I discovered the relation:

sqrt(kT/m) * P = 200.57 AU

This happens to be, to about 1% accuracy, the distance to the circular orbit best fitting my sky survey positions for Barbarossa.

Hi Joe, reading that last post has made me think that the neutrino telescope in the Antarctic is something of a must to approach. It would also be nice to see what the Chinese have come up with from last years solar eclipse.

If we say that light slows on entering a solar system, does the same thing apply for neutrinos? Now I think that the speed of light should be the cube root of the speed of gravity. That would make it about 3.079E 8 metres per second. That would mean that an electron neutrino could go at the speed of light i.e. 2.998E 8 and only have a relativistic increase in mass of about 2.66

let's say we take the reduced mass of an electron 1.449E-31 and divide it by 2.66 to get
5.451E-32 then multiply that by 4.34E-9 the Planck mass. That equals about 2.3656E-40 or about 1.327E-4 eV

Sticking that mass into your equation gives some interesting results, divide by c and we get 1.33 = H0 * P

Not sure what to make of that as yet. One thing to note though, if that mass of an electron neutrino is about right, then its wavelength is going to be in the borderline microwave region, 9.3E-3 metres.

One thing I forgot to mention. If we take that mass of 2.3656E-40 and divide it by 4.34E-9 then we get something that looks like an electron but its down on mass by 2.66 However, the mass units have dropped out, 3.4248E-31 * c^2 will not give us an energy but rather a velocity.

Sloat, Can you, please, post your math statements in math symbols without all the words between the numbers. A / can be used for divide by and = for most of the rest. That way a connection can be seen even by dummies like me between the starting point and the result. Mixing words and numbers confuses me. thanks.

Hi Jim, sorry about that. The problem is, that it's very difficult to write equations on a board. I find myself having to write down equations, just so that I can recognise them.