Re: GR frequency shift formula


vanep@xxxxxxx wrote:
mluttgens@xxxxxxxxxx wrote:
v...@xxxxxxx wrote:
mluttg...@xxxxxxxxxx wrote:
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.

What I posted would be the difference in period due to the position of
the satellite based and Earth based clocks in the gravitational field.
The correction for the GPS would also include a difference in period
based on the relative velocity of the satellite based and Earth based
clocks.

The approximate formula derived from the Schwarzschild solution:

= 1 - [ M/r_sat ] - [ v^2_sat/2 ] + [ M/r_Earth ] + [ v^2_Earth/2 ]

The correction is ~38,500 nanoseconds/ day.

Almost right!

The GR formula gives the approximate prediction correctly. Your formula
is incorrect by over 2000 nanoseconds/day. Using your correction would
result in the GPS being off ~ 615,4 meters/day.

http://www.phys.lsu.edu/mog/mog9/node9.html

My formula, which takes the relative velocity of the satellite based
and Earth based
clocks into account, reduces to
Nu2/Nu1 = 1 + M2*(1/R2-1/d-1/2d)
Shift = Nu2/Nu1 - 1 = 3.66E-10, which is the correct value.
Multiply the shift by the day in nanoseconds (8.64E10), and you get ~
31.6 nanoseconds/day.

The value is 38 thousand 5 hundres nanoseconds/day.

http://www.eftaylor.com/pub/projecta.pdf

Notice that my reduced formula is identical to the GR approximate
formula.

Your formula is garbage. If it was identical to the GR formula it would
make a prediction that was incorrect by over 2000 nanoseconds/day.

My formula is perfectly correct. I introduced d (the orbital radius of
the GPS satellites) = 20200 km, instead of 20200 + the Earth radius =
26578 km.
Then Nu2/Nu1 = 4.45E-10, which , multiplied by the day in nanoseconds,
gives indeed 38.5 microseconds. Sorry!


Using your formula the GPS would have crashed and burned in a couple of
minutes.

Seemingly, you have a problem with simplifying a formula.

You can't even convert your scientific notation to nanoseconds. You
seeminly think your formula predicts a delta of 31.6 nanoseconds/day.





Clearly you must know the GPS is completely functional and the
correction was made using GR so what logic leads you to make the
comment "Clearly, GR is a very poor tool"?

GR can only solve simple cases, like the GPS one.
Remember that to my question
"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 ?",

Folks who understand relativistic physics know what the difference is
between the weak and strong gravitational field. GR certainly is a poor
tool for you because you never read the operations manual. If we were
talking about a chain saw you would have cut off your leg.

What is the limit between those fileds? There is none!
And please don't react so primitively.


Tom rightly answered
"No. But there is no useful approximation that applies, and this is a
quite complicated computation that can only be performed via numerical
simulation.",

whereas my formula at least gives straightforwardly an approximate
solution (Nu2/Nu1=0.9). If you are skeptical, perform the numerical
simulation.

Your formula is wrong. It doesn't give he right answer for the GPS.

My formula gives the correct anwer, see above.

If you could (what I doubt), you should obtain ~0.87.
As long as GRists content themselves with generalities or easy slogans,
they will not be taken seriously.

They will never be takin seriously by cranks such as yourself..

The cranks are those who claim that GR is a wonderful tool, even its
use is very limited.
As the proof of the pudding is in the eating, perform the simulation to
solve my little problem.
But you can't.

Marcel Luttgens.



Marcel Luttgens

.

Relevant Pages

  • Re: GR frequency shift formula
    ... Light emitted from the Sun's surface observed on Earth: ... 'Light emitted from a satellite orbiting at 20200 km from the Earth ... The correction for the GPS would also include a difference in period ... The approximate formula derived from the Schwarzschild solution: ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... Light emitted from the Sun's surface observed on Earth: ... 'Light emitted from a satellite orbiting at 20200 km from the Earth ... the satellite based and Earth based clocks in the gravitational field. ... The correction for the GPS would also include a difference in period ...
    (sci.physics.relativity)
  • Re: GPS corrections
    ... > demands only one equation to explain GPS. ... Rs is the orbital radius of the satellite and Re is the Earth radius ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... Light emitted from the Sun's surface observed on Earth: ... The correction for the GPS would also include a difference in period ... is incorrect by over 2000 nanoseconds/day. ...
    (sci.physics.relativity)
  • Re: GR frequency shift formula
    ... Light emitted from the Sun's surface observed on Earth: ... The correction for the GPS would also include a difference in period ... is incorrect by over 2000 nanoseconds/day. ...
    (sci.physics.relativity)
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