Perihelion precession of Mercury from hyperbolic anisotropy

Main Page | Report this Page

Science Forum Index » Physics - Research Forum » Perihelion precession of Mercury from hyperbolic anisotropy

Page 1 of 1

Author

Message

Just G. Waller

Posted: Tue Feb 27, 2007 12:53 pm

Guest

The hyperbolic anisotropy of vacuum can address
the anomalous precession of any orbiting body, by:

dw = W (Exp(V^2/c^2)- 1),
where
dw is the precession angular velocity,
W is the angular velocity,
V is the magnitude of difference vector of escape velocities
between the orbiting body and observer, and
c = 2.99e+5 km/s is speed of light in the vacuum.

For a quick computation on planet Mercury observed from the Earth,
we only need these data:

GM, standard gravitational parameter of the Sun,
R_m, Mercury average distance to the Sun,
R_e, the Earth average distance to the Sun,
c speed of light in vacuum,

and the favourable alignment: Earth and Mercury distance vectors to
the Sun being orthogonal.

The solar system escape velocity from Mercury, under newtonian laws,
is

V_m = sqrt(2GM/R_m)

The escape velocity from the Earth is

V_e = sqrt(2GM/R_e).

So, it yields a magnitude for the vector difference of

V = sqrt(V_m^2 + V_e^2)=
= sqrt(2GM(1/R_m + 1/R_e)),

because we've assumed each escape velocity is defined in the
direction of its distance vector to the Sun, and they are
orthogonal.

The average angular velocity of Mercury is

W_m = sqrt(GM/R_m^3)

Finally, the precession is

dw = W_m (Exp(V^2/c^2) - 1),
dw = sqrt(GM/R_m^3) (Exp(2GM(1/R_m + 1/R_e)/c^2) - 1)

For average distance of Mercury to the Sun
R_m = 55.0 e+6 km ,
and average distance of the Earth
R_e = 147.0 e+6 km, it yields

dw = 43.12 arcsec per century,
which is a pretty good prediction!.

The factor z = Exp(V^2/c^2) - 1 is actually a shift,
exactly it is a gravitational shift. But, for a body
in an outer orbit with respect to the Earth it would be

z = Exp(-V^2/c^2) - 1
.
So, in this case it is z < 0, and rather than a perihelion precession
we would measure a perihelion delay. The observation of precession
(z > 0)implies a Doppler red-shift for any signal coming from the
body to us, and the observation of an orbit delay implies
a Doppler blue-shift.

The core of this gravitational anisotropy resides in the
difference vector V. For inertial systems, the shift z is
expressed as z = Exp(v/c) - 1, where v is the relative speed
between source and observer. See my last post on this issue at
http://groups.google.com.gi/group/sci.physics.research/browse_thread/thread/48cd5e1ba31c5b28/?hl=en#

For an observer at large distance, such that the gravitational
potential can be regarded as zero, it yields

z = Exp(2GM/rc^2) - 1,
where r is the distance of the body to the center of the system.

Let us compare this solution with the relativistic one for
gravitational
red-shifts, z_s, of non-rotating, uncharged masses which are
spherically
symmetric,

z_s = 1/sqrt(1 - (2GM/rc^2)) - 1

For sake of simplicity, call x = 2GM/rc^2
When we express z and z_s in their respective expansion series,
we see that they match until the second order approximation,

z = x + x^2/2 + x^3/6 + .. + x^n/n! + ...
z_s = x + 3x^2/8 + 5x^3/16 + ...

we see the discrepancy in the second term, it is 4/3. A good
approximation is achieved for low 2GM/r with respect to c^2.

Guest

Posted: Wed Feb 28, 2007 1:24 pm

Just G. Waller <wallermax@hotmail.com> wrote:

[...]

Quote:

The factor z = Exp(V^2/c^2) - 1 is actually a shift,
exactly it is a gravitational shift. But, for a body
in an outer orbit with respect to the Earth it would be

z = Exp(-V^2/c^2) - 1
.
So, in this case it is z < 0, and rather than a perihelion precession
we would measure a perihelion delay.

Then your model is dead. We observe an advance of the perihelion
of Mars, in the amount predicted by general relativity, not a delay.
The same is true for Icarus, whose orbit crosses that of the Earth.
We also observe the preihelion shifts predicted by general relativity
for a number of binary pulsar systems.

Steve Carlip

Ian Parker

Posted: Thu Mar 01, 2007 11:30 am

Guest

The expansion of e^x is of coourse 1 + x + x^2/2! + etc. The first
term provides the correctionfor Mercury but with the wrong sign. In a
weak gravitational field we only have the first term. You need
stronger fields to verify the other terms.

I would also like to point out that GR provides for a number of other
predictions. There are gravitational waves. A pair of pulsars have
been proved to lose energy within 10% of GR. Does this theory predict
gravitons with a spin of 2. I am inclined to feel that either it does
not make any predictions at all, or if it does spin 1 (Maxwell's
equations). Maxwell's Equations produce a loss of energy which is
orders of magnitude too large.

What about black holes. Unusual galaxies emit jets from spinning black
holes. Any alternative theory to GR must correctly predict this.

- Ian Parker

Page 1 of 1 All times are GMT - 5 Hours The time now is Tue Dec 02, 2008 1:40 am Science Forum Index