Sue... wrote:
wrote:
Sue... wrote:
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

Does that mean "Laughing Out Loud -or- Lots of Luck (or Love)"?

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.

Misunderstandings from some GRists! And more generally from crackpots!

Marcel Luttgens


Sue...

dda1 wrote:
wrote:
snipped, the mother****er doesn't understand the Pound Rebka
experiment
Hey, the food is pretty good in your country, why do you persist in
eating **** breakfast, lunch and dinner?

what about not changin tha foken subject lines
yo stooped foken nazimothofaka

Sue... wrote:
wrote:
Sue... wrote:
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

wrote in message oups.com...

Sue... wrote:
wrote:
Sue... wrote:
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.

As false as the The Lorentz transformation (LT), by M. Luttgens?
http://perso.wanadoo.fr/mluttgens/LTfalse.htm

Dirk Vdm

wrote:
Sue... wrote:
wrote:
Sue... wrote:
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.

Equation 14 looks reasonable:
http://arxiv.org/abs/gr-qc/9606079

It is in terms of gravitational potential so
would predict the lowest frequency at the surface
then increasing as the clock either moved up or down.
That is consistant with free pendulms, long know to
decrease in frequency when moved up or down from the
surface.

As for your equation above, Nu1 should equal Nu2.
....unless you have a mechanism to create or destroy
the news of events as it is propagated. Maybe that
is what is happening on a noisy satellite feed when
the announcer freezes for a few seconds? )

Sue...




Marcel Luttgens

dda1 wrote:
my mother wrote:
Stop ****ing your mother and eating your own ****. It makes you and
your offspring imbeciles. Wait, you are already and imbecile!

this is a wonderful think, informin tha entire
population about tha writings of your mother

who wold even think such a thing could
happen this place, these peoples are psykos

doin his own mother, she tellin him ta stop
doin it

disgustin, but thank yo anyways

Sue... wrote:
wrote:
Sue... wrote:
wrote:
Sue... wrote:
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.

Equation 14 looks reasonable:
http://arxiv.org/abs/gr-qc/9606079

Equation 14 should here be written in terms of the shaft depth!
Perhaps a true GR specialist could do this.


It is in terms of gravitational potential so
would predict the lowest frequency at the surface
then increasing as the clock either moved up or down.
That is consistant with free pendulms, long know to
decrease in frequency when moved up or down from the
surface.

As for your equation above, Nu1 should equal Nu2.

Look at my equation: the shift is given by (GMe/2Re^3c^2) * (Re^2-d^2)
Only if d = Re is Nu2 equal to Nu1 (the depth of the shaft is 0).
If d = 0 (the receiver is at the Earth's center), the shift becomes
GMe/2Re*c^2.
This shift should rather easily be obtained with GR.

...unless you have a mechanism to create or destroy
the news of events as it is propagated.

?????

Marcel Luttgens

Maybe that
is what is happening on a noisy satellite feed when
the announcer freezes for a few seconds? )

Sue...




Marcel Luttgens

wrote in message ups.com...

Dirk Van de moortel wrote:
wrote in message oups.com...

Sue... wrote:
wrote:
Sue... wrote:
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.

As false as the The Lorentz transformation (LT), by M. Luttgens?
http://perso.wanadoo.fr/mluttgens/LTfalse.htm

Give the correct equation if mine is false! But you can only parrot!

You can't handle the meaning of the variables in equations, Marcel:
http://users.telenet.be/vdmoortel/di...idntUseSR.html
For your own sake, try to avoid them, as long as you still can.

Dirk Vdm

wrote:
Sue... wrote:
wrote:
Sue... wrote:
wrote:
Sue... wrote:
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.

Equation 14 looks reasonable:
http://arxiv.org/abs/gr-qc/9606079

Equation 14 should here be written in terms of the shaft depth!
Perhaps a true GR specialist could do this.

Perhaps they anticipated it might be used on another
planet where the shaft depth would be meaningless.



It is in terms of gravitational potential so
would predict the lowest frequency at the surface
then increasing as the clock either moved up or down.
That is consistant with free pendulms, long know to
decrease in frequency when moved up or down from the
surface.

As for your equation above, Nu1 should equal Nu2.

Look at my equation: the shift is given by (GMe/2Re^3c^2) * (Re^2-d^2)
Only if d = Re is Nu2 equal to Nu1 (the depth of the shaft is 0).
If d = 0 (the receiver is at the Earth's center), the shift becomes
GMe/2Re*c^2.
This shift should rather easily be obtained with GR.

I would rather review the arithmetic for the bellhop paradox
again than your equations after your statement that light
changes frequency absent a change in path length.

Are you saying if the paper substitutes your equation
for the one they used then it will make the paper wrong?
If so then I agree with you because the paper appears
correct as it is.... to degee of showing the spacetime
interval tau instead of time, a detail many writers overlook.

Sue...




...unless you have a mechanism to create or destroy
the news of events as it is propagated.

?????

Marcel Luttgens

Maybe that
is what is happening on a noisy satellite feed when
the announcer freezes for a few seconds? )

Sue...




Marcel Luttgens

Sue... wrote:
wrote:
Sue... wrote:
wrote:
Sue... wrote:
wrote:
Sue... wrote:
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.

Equation 14 looks reasonable:
http://arxiv.org/abs/gr-qc/9606079

Equation 14 should here be written in terms of the shaft depth!
Perhaps a true GR specialist could do this.


"Perhaps they anticipated it might be used on another
planet where the shaft depth would be meaningless."

Ha Ha !



It is in terms of gravitational potential so
would predict the lowest frequency at the surface
then increasing as the clock either moved up or down.
That is consistant with free pendulms, long know to
decrease in frequency when moved up or down from the
surface.

As for your equation above, Nu1 should equal Nu2.

Look at my equation: the shift is given by (GMe/2Re^3c^2) * (Re^2-d^2)
Only if d = Re is Nu2 equal to Nu1 (the depth of the shaft is 0).
If d = 0 (the receiver is at the Earth's center), the shift becomes
GMe/2Re*c^2.
This shift should rather easily be obtained with GR.

I would rather review the arithmetic for the bellhop paradox
again than your equations after your statement that light
changes frequency absent a change in path length.

"Are you saying if the paper substitutes your equation
or the one they used then it will make the paper wrong?"

As long as their equation 14 is not written in terms of Me, Re and d,
I prefer to wait and see.

A nicer form of the "shaft" shift formula is
shift = GMe/2Rec^2 * (1-d^2/Re^2),
where d is the distance between the bottom of the shaft and the
Earth's center.
The shift could of course be expressed in terms of the
shaft depth = Re-d. Replacing d by Re-depth, one gets
shift = GMe/2Rec^2 * (1-(Re-depth)^2/Re^2)

If the depth = Re, one is left with shift = GMe/2Rec^2 (1)
Let's remember that the signal is sent from the Earth's surface
to the bottom of the shaft.

Notice that a signal emitted at infinity towards the Earth's surface
would be observed to be shifted by GMe/Rec^2 (also according to GR).
The fact that this shift is twice the "shaft" shift (1) must be
more than a coincidence!

Marcel Luttgens

If so then I agree with you because the paper appears
correct as it is.... to degee of showing the spacetime
interval tau instead of time, a detail many writers overlook.

Sue...




...unless you have a mechanism to create or destroy
the news of events as it is propagated.

?????

Marcel Luttgens

Maybe that
is what is happening on a noisy satellite feed when
the announcer freezes for a few seconds? )

Sue...




Marcel Luttgens