Re: GR frequency shift formula




mluttgens@xxxxxxxxxx wrote:
vanep@xxxxxxx wrote:
mluttgens@xxxxxxxxxx wrote:
vanep@xxxxxxx wrote:
Tom Roberts wrote:
vanep@xxxxxxx wrote:
Clocks tick at different rates due to their position in the
gravitational field and the total photon energy remains constant as it
changes position in the gravitational field.

I repeat: that is _one_ way to interpret the equations of GR. But it is
not the only one. Both rate and energy are coordinate-dependent
quantities and you are implicitly assuming a specific coordinate system
here; use different coordinates and you can get different results for
clock tick rates and "photon" energy.

I don't dispute that. The 'loss of energy' interpretation can lead to
misrepresentation such as Marcel made. My impression of Marcels
comments is he might conclude that GR predicts proper clock rates are
coordinate-dependent.



Koobee Wublee rightly wrote on April 6, 2006:

"GR's best traits are

** To wait for an actual observation to occur and from its portfolio
of variety of explanations to produce the one that taylor-fits this
particular observation

** To subjectively interpret the data to fit one from GR's impressive
portfolio of possible predictions of the same event."

You are about as ignorant as one can be. Read the literature then
actually learn something about GR.

Moreover, GR is simply not applicable to the Earth interior.

How would you know? Everything you've said in this thread proofs you
know nothing about relativistic physics. That's not a sin in itself,
but complaining about physics you don't understand, havn't read the
literature for, and spent a noticable amount of effort ignorantly
trying to debunk, is a sin against intellectual honesty.

Bruce

Till now, I didn't get a satisfactory GR formula for the Earth
interior.

You didn't explain the meaning of r in your formula
dTau = [ 1+ (1/2)(Mr^2/R^3) - (3/2)(M/R) ] dt ,which, btw,
is different from the formula given by Tom Roberts
(f_det = sqrt((1+2P_detector/c2)/(1+2P_source/c2)) f_source),
that doesn't apply to the interior of the Earth.

Where is your correct GR formula giving the frequency Nu1 at the Earth
surface
of a light of frequency Nu0 emitted from the Earth center?

The approximate GR formula is the same as the one I presented before,
i.e.
Nu1 / Nu0 = 1 + GM/2Rc^2, where M and R are respectively the mass and
the radius
of the Earth (It is assumed that the Earth is homogeneous, transparent
and not rotating).
Notice that I then didn't use GR at all to get such formula.

The GR formula is Nu1 / Nu0 = (3X - 1) / 2X, where
X = sqrt(1 - 2GM/Rc^2).

So, I have to admit that GR can solve such little problem. :-)

Otoh, it is practically useless for more complicated situations, for
instance:

"Two non-rotating spherical celestial objects 1 and 2 are orbiting
along
their common center of gravity.
The masses and radii of the objects are respectively M1, R1 and M2,R2.
The distance between centers is d.
Object 1 emits a light of frequency Nu1. What is the frequency Nu2
of the light received by an observer situated on object 2 at a distance

d - (R1+R2) from the emitter? ",

whereas an approach via the potential enegy of photons of mass E/c^2
straightforwardly leads to a solution.

Marcel Luttgens


Marcel Luttgens


In particular if you use coordinates local to source and others local to
detector, both determined by standard clocks and rulers, you will
conclude that clocks tick at the same rate and "photons" gain or lose
energy/frequency. In some sense these are more natural coordinates than
the ones you assumed for your statement above (your coordinate clocks
are not standard).

I ignore the quantum aspects of photons; consider light rays
instead and measure their frequencies.


Tom Roberts tjroberts@xxxxxxxxxx

.



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