Re: Einstein interpretation of gravitational redshift is misleading


Sue... wrote:
mluttgens@xxxxxxxxxx wrote:
Sue... wrote:
mluttgens@xxxxxxxxxx wrote:


As the receiver is moving relatively to the source, why doesn't he
observe a blue shift due to the Doppler effect, instead of no shift
at all?


The Mossbauer 'receiver' is extremely narrow due to the high
Q of the atomic oscillator. It only observes (absorbs) light
which is of the right frequency and phase to permit an
efficient transition to a permissible energy level.

A wide band receiver will intercept the transitions at a
higher rate due to the closing motion, just as you have
described.

Sue...

Thank you.

Only crackpots would stick to the explanation according to which
"clocks which run the faster the higher they are located in the
potential, whereas the energy and frequency of the propagating photon
do not change with height."

But, nevertheless, most GRists will probably not openly recognize that
such explanation is "misleading".

Marcel Luttgens

LOL
I am looking for a Pound-Rebka-Snider experiment down a
mine shaft. I think that would clear up a lot of the misunderstanding
but so far I haven't found one. Either my research skills are
getting rusty or graduate students have a softer life these days.

Sue...

You wrote:

"I am looking for a Pound-Rebka-Snider experiment down a
mine shaft. I think that would clear up a lot of the misunderstanding
but so far I haven't found one."

In the meantime, you could content yourself with the formula easily
obtained by using the potential energy of the photons:

Nu1 is the frequency of the signal sent to the bottom
of the shaft.
Nu2 is the frequency of the signal received at the bottom.
d is the distance of the bottom of the shaft to the Earth's center.
Me and Re are respectively the mass and the radius of the Earth.
The shift Nu2/Nu1 - 1 = (GMe/2Re^3c^2) * (Re^2-d^2)

If GR doesn't get the same formula, it is false.

Marcel Luttgens

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