Re: GR frequency shift formula


vanep@xxxxxxx wrote:
mluttgens@xxxxxxxxxx wrote:
Tom Roberts wrote:
mluttgens@xxxxxxxxxx wrote:
> Do I have to conclude that GR cannot numerically solve this elementary
> problem where
> M1 = M* ( M* = 1 solar mass), R1 = 5 Km, M2 = 5 M*, R2 = 20 km, d = 50
> km ?

No. But there is no useful approximation that applies, and this is a
quite complicated computation that can only be performed via numerical
simulation. You do realize those are neutron-star densities, don't you?

AFAIK there are no observed binary systems involving compact objects
with anything close to that small a distance, probably because the
system will decay into a collision in a short time on astronomical scales.


Are you kidding? The approximate formula
Nu2/Nu1 = 1 - M1*(1/R1-1/(d-R2)-1/(2*d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
applies to all cases, even the theoretical ones.

Let's apply it to two well known cases:

1) Light emitted from the Sun's surface observed on Earth:

M2 = 0 'The Earth mass is negligible
R1 = 6.95E+10 'Sun radius in cm
R2 = 6.38E+08 'Earth radius in cm
d = 1.5E+13 'Mean Earth-Sun distance

The formula reduces to
Nu2/Nu1 = 1 - M1*(1/R1-1/(d-R2)-1/(2*d))
Shift = Nu2/Nu1 - 1 = -2.11E-6

2) 'Light emitted from a satellite orbiting at 20200 km from the Earth
center

M1 = 0 'The satellite mass is negligible
R1 = 0 'The satellite radius is negligible
M2 = 5.977E+27 'Mass of the Earth in g
M2 = G * M2 / c ^ 2
R2 = 6.38E+08 'Radius of the Earth in cm
d = 2.02E+09 'cm

The formula reduces to
Nu2/Nu1 = 1 + M2*(1/R2-1/d-1/2d)
Shift = Nu2/Nu1 - 1 = 3.66E-10

But I agree, you can't get the formula with GR.

Marcel Luttgens



Tom Roberts tjroberts@xxxxxxxxxx

As Tom points out you can derive a weak field approximation. For
example [2]: approximating the Earth as spherically symmetric and
non-rotating we can derive the approximate formula from the
Schwarzschild solution. In geometric units

M_Earth = .00444 meters
r_satellite = 26,571,000 meters
r_Earth = 6,371,000 meters

= 1 - [ M_Earth / r_satellite ] + [ M_Earth / r_Earth ] = 1. 5.2981E-10

Sorry, your formula is false. I gave the correct solution, which
corresponds to the 31.6 microseconds/day that are taken into account to
correct the clocks of the GPS satellites (whose orbital radius is 20200
km, as you probably know).
For such satellites, the GR approximate formula is
dt(s)/dt(e) = 1 + M*(1/Re - 3/2Rs) = 1 + shift, where

M = 4.44E-3 G*Mearth/c^2 in m
day = 8.64E+10 in microseconds
Re = 6.371E+6 Earth radius in m
Rs = 2.02E+7 Satellite orbital radius


Without checking your formula, I would suspect, the wrong answer
originates with concluding the difference between r_satellite and
r_Earth can be ignored. Tom is certainly correct about the difficulty
associate with a strong field calculation based on the reasons he gave.

Which reasons? That close binaries are unstable? Tom should have
considered the mathematical aspect of the problem, and not prefer
escapism.
Clearly, GR is a very poor tool.

Marcel Luttgens

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