What's up with gravity wave detection?

(Eric Flesch) wrote in message ...
On Fri, 20 Aug 2004 17:34:31 +0000 (UTC),
wrote:
In sci.astro Eric Flesch wrote:


Of course not. Bosons classically manifest routinely and so their
gravitational vectors can be described. But photons have *no*
classical manifestation between emission and absorption. There is a
real difference here in their behavior "in the wild".

"Observing" them destroys them; the eye/receiver converts them into a
different form of energy. It can be deduced where they are, and what
has occured, "in the wild". A photon emitted at A towards B, which is
at distance 2 light seconds, can be confidently assumed to be at the
halfway point after exactly 1/2 sec- and a receiver placed at that
point at exactly that instant will show so.
As for the gravitational effect on the velocity of a photon, that is
seen routinely when a star passes behind the sun- the path of the
photon is changed by the sun's gravity (per Newton)

Just trap a bunch
of radiation in a mirrored box and glue it next to a mass. The
radiation will be attracted toward the mass

No it won't be. The mass might have some miniscule effect on the
local geometry, that is all.

Are you serious? You claim that a mirrored box filled with radiation does
not weight more than an empty box?

Would you like to describe how a mirror works? How efficient mirrors
are? Every mirror that we've ever built, if we build the box that you
describe, toss in some photons and close the box, the photons would be
absorbed (and converted to heat) in an infinitesimal fraction of a
second. Are you resorting to a "perfect" mirror of some kind? Would
this work the same as standard mirrors (in which photons are absorbed
& re-emitted with some efficiency) or do you prescribe a perfect
mirror which actually re-directs the same photon in some
perpetual-motion kind of way? Your illustration is fanciful. No
actual experiment has ever "weighed" in-flight radiation.

As above, the effect of gravitation on EMR is readily observable.

Here's a simple exercise. I assume you accept that electromagnetic radiation
is red-shifted in a gravitational field, right?

It depends on the vector. Photons departing the gravitational field
are red-shifted, certainly.

So consider a box with a
mirror at the top and one at the bottom, containing radiation in a coherent
state (you accept QED, right?) with an expectation value of momentum that's
in the purely vertical direction and an expectation value of wave packet
width (z^2-z^2) that's small compared to the size of the box. Compute
the momentum transfer to the mirrors, using however much QED you like. You
will find that the momentum transfer to the bottom, where the energy is
blue-shifted, is greater than the momentum transfer to the top. That means
``the radiation will be attracted toward the mass.''

In that sense you are right. But I am reminded of one of Bohr's
refutations of Einstein's gedankenexperimenten where Einstein's
premise of increased mass was countered by the different rates of time
flow in a differential gravitational field. I suspect the same
applies here. In any event, this does not show that the "in-flight"
photon exerts gravitation.

This would suggest, that for EMR, gravity is a one way street: the sun
exerts an influence on passing photons, but you claim that they do not
reciprocate---
highly unlikely

Jim G
c'=c=v

"Androcles" wrote in message ...
"Jim Greenfield" wrote in message
om...
| Paul Lawler wrote in message
.125.202...
| "Androcles" wrote in news:TmuVc.3953
| :
|
|
| Risking showing my naivitie, what is REALLY being discussed- a "one
| of" change in gravitational field strength (pulse/wave), or a SERIES
| of equal strength waves emanating from a static body?
|
| Jim G
| c'=c+v

Jim, this discussion is not about the existence or non-existence of gravity
waves, but about their amplitude being great enough to be detectable. Simply
spinning the Earth in the lunar gravity produces tides and when we include
solar gravity we have neap and spring tides. If the lunar orbit were highly
elliptical we'd have higher tides at perigee than at apogee. Thus we would
have a detectable gravity 'wave'; they do exist, and can be detected.

Yep. As I told Old Man, I weigh less with moon above than 12 hours
later. So the moon produces a gravity wave of frequency 1/24hour,
right?

LIGO, however, is about detecting a gravitational field from a supernova at
a distance of a kiloparsec = 3260 light years, where some quantity of matter
is
completely converted to energy (E= mc^2) and the resultant gravity field is
reduced. That would be a step pulse.
Or it could be the field from a pulsar in orbit about a neighbour that is
periodically approaching and receding from us, and that would be a
sinusoidal wave. So the answer to your question is : both. However, the
supernova (which may produce a pulsar as a remnant) is the greater.

If a star explodes, the "center of gravity" of that star remains in
the same place afterward. As I intuitively feel that the particles
which comprise EMR DO exert gravitational force themselves, therefore
no pulse/wave, as the star still "acts" the same after exploding.
However, as the EMR dissipates, opening up the angle from us (from a
point to an expanding cloud), there should be a gradual decline in
field strength towards/from that center of gravity. Undetectable
change until the outer ring of the burst is over a significant arc to
us = no wave (detectable = Ligo wont work for Sn

If you want to express the problem mathematically: let delta be the smallest
amplitude detectable by the instrument used.
Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude at r
where the instrument is placed be A/r^2 = delta.
Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta and
is not detectable.
LIGO has a real delta, so from that estimate the greatest imaginable A and
calculate
A/r^2 = delta
r^2/A = 1/delta
r = sqrt(A/delta)

Androcles.

Don't math me :-(

Jim G
c'=c+v

On Mon, 23 Aug 2004 07:36:10 GMT, (Eric Flesch) wrote:

On Fri, 20 Aug 2004 17:34:31 +0000 (UTC),
wrote:
In sci.astro Eric Flesch wrote:
You're missing the point. The stress-energy tensor is a classical
description which assumes continuous motion. But QED shows that the
photon path is the summation of all possible paths (diffraction
gratings are an application of this) and the delayed-choice experiment
shows explicitly that the travelling photon cannot be pinpointed to
any particular location in its presumed path(s). The point is that
the "travelling photon" can be modelled only by a quantum description,
and the classical stress-energy tensor does not apply.

Everything you say here is just as true of neutrons, or protons, or
any other elementary particle. Are you saying that we should therefore
not use GR at all?

Of course not. Bosons classically manifest routinely and so their
gravitational vectors can be described. But photons have *no*
classical manifestation between emission and absorption. There is a
real difference here in their behavior "in the wild".

What is a `classical manifestation' of gravity ?

... You can write down a low-energy effective action for
the gravitational field without knowing details of quantum gravity.

"Details"? Quantum gravity is vaporware. You can't quote it as
though it were a reference.

Huh?

You can't have it both ways here. If you want to talk about what ``QED
shows,'' you have to allow standard techniques from QED.

You mystify me. What have I said to limit the application of QED?
I'm saying that where QED's description differs from GR, QED rules.
GR is not, and was never meant to be, a description of the nature of
light.

Then why are photons `described?

Just trap a bunch
of radiation in a mirrored box and glue it next to a mass. The
radiation will be attracted toward the mass

You're kidding?

No it won't be. The mass might have some miniscule effect on the
local geometry, that is all.

Are you serious? You claim that a mirrored box filled with radiation does
not weight more than an empty box?

So radiation (photons ) are affected by gravity? Aren't we going
around in circles?

Would you like to describe how a mirror works? How efficient mirrors
are? Every mirror that we've ever built, if we build the box that you
describe, toss in some photons and close the box, the photons would be
absorbed (and converted to heat) in an infinitesimal fraction of a
second. Are you resorting to a "perfect" mirror of some kind? Would
this work the same as standard mirrors (in which photons are absorbed
& re-emitted with some efficiency) or do you prescribe a perfect
mirror which actually re-directs the same photon in some
perpetual-motion kind of way? Your illustration is fanciful. No
actual experiment has ever "weighed" in-flight radiation.

Some would be absorbed, some deflected, and some escape without
interaction.

Here's a simple exercise. I assume you accept that electromagnetic radiation
is red-shifted in a gravitational field, right?

It depends on the vector. Photons departing the gravitational field
are red-shifted, certainly.

So consider a box with a
mirror at the top and one at the bottom, containing radiation in a coherent
state (you accept QED, right?) with an expectation value of momentum that's
in the purely vertical direction and an expectation value of wave packet
width (z^2-z^2) that's small compared to the size of the box. Compute
the momentum transfer to the mirrors, using however much QED you like. You
will find that the momentum transfer to the bottom, where the energy is
blue-shifted, is greater than the momentum transfer to the top. That means
``the radiation will be attracted toward the mass.''

What is difference of mass between the mirrors? What is `top' and what
`bottom'? How does the `radiation' which doesn't respond to gravity,
decide?

In that sense you are right. But I am reminded of one of Bohr's
refutations of Einstein's gedankenexperimenten where Einstein's
premise of increased mass was countered by the different rates of time
flow in a differential gravitational field. I suspect the same
applies here. In any event, this does not show that the "in-flight"
photon exerts gravitation.

Eric

On 23 Aug 2004 17:36:06 -0700, (Jim
Greenfield) wrote:
As for the gravitational effect on the velocity of a photon, that is
seen routinely when a star passes behind the sun- the path of the
photon is changed by the sun's gravity (per Newton)

You're confusing topics; we were discussing "in-flight" photons
*exerting* gravity.

In that sense you are right. But I am reminded of one of Bohr's
refutations of Einstein's gedankenexperimenten where Einstein's
premise of increased mass was countered by the different rates of time
flow in a differential gravitational field. I suspect the same
applies here. In any event, this does not show that the "in-flight"
photon exerts gravitation.

This would suggest, that for EMR, gravity is a one way street: the sun
exerts an influence on passing photons, but you claim that they do not
reciprocate---
highly unlikely

Why, because you say so?

Eric

"Jim Greenfield" wrote in message
m...
| "Androcles" wrote in message
...
| "Jim Greenfield" wrote in message
| om...
| | Paul Lawler wrote in message
| .125.202...
| | "Androcles" wrote in
news:TmuVc.3953
| | :
| |
| |
| | Risking showing my naivitie, what is REALLY being discussed- a "one
| | of" change in gravitational field strength (pulse/wave), or a SERIES
| | of equal strength waves emanating from a static body?
| |
| | Jim G
| | c'=c+v
|
| Jim, this discussion is not about the existence or non-existence of
gravity
| waves, but about their amplitude being great enough to be detectable.
Simply
| spinning the Earth in the lunar gravity produces tides and when we
include
| solar gravity we have neap and spring tides. If the lunar orbit were
highly
| elliptical we'd have higher tides at perigee than at apogee. Thus we
would
| have a detectable gravity 'wave'; they do exist, and can be detected.
|
| Yep. As I told Old Man, I weigh less with moon above than 12 hours
| later. So the moon produces a gravity wave of frequency 1/24hour,
| right?
Relatively speaking, yes. Of course it is the spin of the Earth that
actually changes your position and produces the wave, you are simply moving
in a fixed field, the strength of which is a function of distance. 1/24hour
may be misconstrued, though. 1 cycle every 24 hours is not the same as 1
cycle in 2 minutes and 30 seconds, which is 1/24th of an hour. To be a
little more precise, the moon orbits the Earth 13 times a year (but not
exactly) and the Earth orbits the sun. 24 hours is the time from noon to noo
n (sun at zenith) but a distant star moves about 4 minutes a day from
midnight to midnight. Which star is overhead at midnight depends on your
longitude. But rougly speaking, the gravity wave has a frequency of one
cycle per day.



|
| LIGO, however, is about detecting a gravitational field from a supernova
at
| a distance of a kiloparsec = 3260 light years, where some quantity of
matter
| is
| completely converted to energy (E= mc^2) and the resultant gravity field
is
| reduced. That would be a step pulse.
| Or it could be the field from a pulsar in orbit about a neighbour that
is
| periodically approaching and receding from us, and that would be a
| sinusoidal wave. So the answer to your question is : both. However, the
| supernova (which may produce a pulsar as a remnant) is the greater.
|
| If a star explodes, the "center of gravity" of that star remains in
| the same place afterward.

Yes. There may be a shell of matter that leaves the star, but momentum is
conserved. The same is true for a rocket. We see the rocket accelerate, but
the exhaust is flying away in the opposite direction and the centre of mass
of the combined exhaust and rocket only moves with its original velocity
before the engine was fired. The combined momentum of the entire Universe is
zero.

| As I intuitively feel that the particles
| which comprise EMR DO exert gravitational force themselves, therefore
| no pulse/wave, as the star still "acts" the same after exploding.
| However, as the EMR dissipates, opening up the angle from us (from a
| point to an expanding cloud), there should be a gradual decline in
| field strength towards/from that center of gravity. Undetectable
| change until the outer ring of the burst is over a significant arc to
| us = no wave (detectable = Ligo wont work for Sn

Intuition is a dangerous tool. I don't recommend it. Better to prove a
theorem mathematically and then see if intuition agrees. Thunder and
lightning arrive at different times, and a child's intuition is that they
are seperate events. An adult sees it differently. Until Copernicus,
intuition told us the Earth is at the centre if the universe. After all, we
see the sun cross the sky daily, it MUST be going around us. With greater
knowledge we revise our view that we are turning toward and away from the
sun. Never trust intuition, it is bane of science and the boon of religion.


|
| If you want to express the problem mathematically: let delta be the
smallest
| amplitude detectable by the instrument used.
| Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude
at r
| where the instrument is placed be A/r^2 = delta.
| Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta
and
| is not detectable.
| LIGO has a real delta, so from that estimate the greatest imaginable A
and
| calculate
| A/r^2 = delta
| r^2/A = 1/delta
| r = sqrt(A/delta)
|
| Androcles.
|
| Don't math me :-(
|
| Jim G
| c'=c+v -- don't math me. :-)
Androcles

"Androcles" wrote in message ...
"Jim Greenfield" wrote in message
m...
| "Androcles" wrote in message
...
| "Jim Greenfield" wrote in message
| om...
| | Paul Lawler wrote in message
.125.202...
| | "Androcles" wrote in
news:TmuVc.3953
| | :
| |
| |
| | Risking showing my naivitie, what is REALLY being discussed- a "one
| | of" change in gravitational field strength (pulse/wave), or a SERIES
| | of equal strength waves emanating from a static body?
| |
| | Jim G
| | c'=c+v
|
| Jim, this discussion is not about the existence or non-existence of
gravity
| waves, but about their amplitude being great enough to be detectable.
Simply
| spinning the Earth in the lunar gravity produces tides and when we
include
| solar gravity we have neap and spring tides. If the lunar orbit were
highly
| elliptical we'd have higher tides at perigee than at apogee. Thus we
would
| have a detectable gravity 'wave'; they do exist, and can be detected.
|
| Yep. As I told Old Man, I weigh less with moon above than 12 hours
| later. So the moon produces a gravity wave of frequency 1/24hour,
| right?
Relatively speaking, yes. Of course it is the spin of the Earth that
actually changes your position and produces the wave, you are simply moving
in a fixed field, the strength of which is a function of distance. 1/24hour
may be misconstrued, though. 1 cycle every 24 hours is not the same as 1
cycle in 2 minutes and 30 seconds, which is 1/24th of an hour. To be a
little more precise, the moon orbits the Earth 13 times a year (but not
exactly) and the Earth orbits the sun. 24 hours is the time from noon to noo
n (sun at zenith) but a distant star moves about 4 minutes a day from
midnight to midnight. Which star is overhead at midnight depends on your
longitude. But rougly speaking, the gravity wave has a frequency of one
cycle per day.


More precise !!!!,how for goodness sake did you lot manage to get
control of astronomy I will never know but the trail leads back to
Newton.

Poor Isaac,gave you a system that competes with the Tychonian
quasi-geocentric system based on Flamsteed.With concentration now
turning to Copernicus and the attempt by Brahe to return to a
quasi-geocentric view,it is only a matter of time before Newton's
alternative version rears its ugly head.

"PH?NOMENON IV.
That the fixed stars being at rest, the periodic times of the five
primary planets, and (whether of the sun about the earth, or) of the
earth about the sun, are in the sesquiplicate proportion of their mean
distances from the sun.

http://members.tripod.com/~gravitee/phaenomena.htm

This is how that statement works out graphically -


http://www.absolutebeginnersastronomy.com/sidereal.gif




Even if you ever get round to reading 'The Book Nobody Read' you are
unlikely to pick up on the idea of heliocentric motion taken from the
center of the Earth's orbit rather than the Sun's center,poor Isaac in
the best tradition of mathematicians fudged what he could not
understand.

Pity you can't read astronomical language and pity it took me a long
time to understand that,the ability is a gift rather than acquired.The
recent recovery of a more accurate history of investigative techniques
in geology, astronomy,clockmaking and multiple other avenue paves the
way for setting off the awful mistake Newton made in adopting his
particular nasty quasi-geocentric outlook.

You live your life living with Newton's personal revenge but defiance
until death is sure a lousy reward for anyone,aetherist,relativist and
bottom line.Simply stated,Newton got it wrong and in a very
fundamental way via Flamsteed.

http://hypertextbook.com/facts/1999/JennyChen.shtml

That is how close your entire discipline is to oblivion.



|
| LIGO, however, is about detecting a gravitational field from a supernova
at
| a distance of a kiloparsec = 3260 light years, where some quantity of
matter
| is
| completely converted to energy (E= mc^2) and the resultant gravity field
is
| reduced. That would be a step pulse.
| Or it could be the field from a pulsar in orbit about a neighbour that
is
| periodically approaching and receding from us, and that would be a
| sinusoidal wave. So the answer to your question is : both. However, the
| supernova (which may produce a pulsar as a remnant) is the greater.
|
| If a star explodes, the "center of gravity" of that star remains in
| the same place afterward.

Yes. There may be a shell of matter that leaves the star, but momentum is
conserved. The same is true for a rocket. We see the rocket accelerate, but
the exhaust is flying away in the opposite direction and the centre of mass
of the combined exhaust and rocket only moves with its original velocity
before the engine was fired. The combined momentum of the entire Universe is
zero.

| As I intuitively feel that the particles
| which comprise EMR DO exert gravitational force themselves, therefore
| no pulse/wave, as the star still "acts" the same after exploding.
| However, as the EMR dissipates, opening up the angle from us (from a
| point to an expanding cloud), there should be a gradual decline in
| field strength towards/from that center of gravity. Undetectable
| change until the outer ring of the burst is over a significant arc to
| us = no wave (detectable = Ligo wont work for Sn

Intuition is a dangerous tool. I don't recommend it. Better to prove a
theorem mathematically and then see if intuition agrees. Thunder and
lightning arrive at different times, and a child's intuition is that they
are seperate events. An adult sees it differently. Until Copernicus,
intuition told us the Earth is at the centre if the universe. After all, we
see the sun cross the sky daily, it MUST be going around us. With greater
knowledge we revise our view that we are turning toward and away from the
sun. Never trust intuition, it is bane of science and the boon of religion.


|
| If you want to express the problem mathematically: let delta be the
smallest
| amplitude detectable by the instrument used.
| Let a pulse (or wave) of amplitude A be emitted at 0 and the amplitude
at r
| where the instrument is placed be A/r^2 = delta.
| Then the amplitude at A/(r+epsilon)^2 (epsilon 0) is less than delta
and
| is not detectable.
| LIGO has a real delta, so from that estimate the greatest imaginable A
and
| calculate
| A/r^2 = delta
| r^2/A = 1/delta
| r = sqrt(A/delta)
|
| Androcles.
|
| Don't math me :-(
|
| Jim G
| c'=c+v -- don't math me. :-)
Androcles

On Tue, 24 Aug 2004 05:46:06 GMT, (Eric Flesch) wrote:

On 23 Aug 2004 17:36:06 -0700, (Jim
Greenfield) wrote:

.....

But...but... shouldn't we find out what `gravity' really is before we
go too much farther. Newton - a force, Einstein - an alteration in
spacetime around a mass. But why does one mass move toward another
mass against its own inertia. If it is `falling into another masses
spacetime `hole' - why? Don't say it's a force because then we are
back to Newton, who admitted he didn't know why the observation was
valid. Just a simple answer if possible: what is gravity? We have
all observed the action... now - why? what is a `graviton'? what is a
`gravitational wave'? what is quantum gravity?

* dkomo:
I found an old PBS documentary on VHS from 1991 called _The Astronomers_
at the local public library. One of the programs in the series was
"Waves of the Future" about gravitational waves. In the program Kip
Thorne was shown making a bet with one of his collaborators on gravity
wave theory that these waves would positively be detected by 2000.

I found this both humorous and a touch sad. The program described some
of the early planning for LIGO (Laser Interferometer Gravitational Wave
Observatory). Curious, I went to the LIGO web site to see what was
going on. I found nothing of substance there -- just a lot of slick PR.

So my question is, what are the prospects that gravity waves will be
detected anytime soon? Is LIGO still having technical problems or what?
It is now 2004, after all. Other detection labs are being built around
the world. Are these labs going to have any better luck?

From the chronology availble at the web-site, url:
http://www.ligo.caltech.edu/LIGO_web/PR/scripts/chrono.html, it seems
nothing has happened after August 2002 when the first "scientific operation
of all three interferometers in S1" was run (note: without me knowing
anything about it I'd say 3 interferometers are the least number needed to
establish any kind of direction for a short-duration signal, but then it
would still be open which of the two halves of the sky it came from).

I remember SciAm had an article essentially saying between the lines that
the noise levels are too high to expect any useful (non-ambigous) result.

So a _detection_ would probably be a major event, contradicting theory...



Also, what are people's opinions about gravity waves? Is it possible
that these are a scientific dead end like the decay of the proton turned
out to be?

Anything is possible, but gravity waves are associated with some extremely
precise predictions of increasing rotation rate of pulsars. Hulse and
Taylor got the Nobel Prize for that work in 1993. So it seems gravity waves
are real, or else the universe is playing us Yet Another magician's joke!


If gravity waves are never detected, what are the
implications for the general theory of relativity?

Serious.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

Intuition is a dangerous tool. I don't recommend it. Better to prove a
theorem mathematically and then see if intuition agrees. Thunder and
lightning arrive at different times, and a child's intuition is that
they are seperate events. An adult sees it differently. Until
Copernicus, intuition told us the Earth is at the centre if the
universe. After all, we see the sun cross the sky daily, it MUST be
going around us. With greater knowledge we revise our view that we
are turning toward and away from the sun. Never trust intuition, it
is bane of science and the boon of religion.

But aren't you using intuition to discard relativistic addition of
velocities in your c'=c+v (or is that c=c'+v?).


daveL

"Dave" wrote in message
...
|
| Intuition is a dangerous tool. I don't recommend it. Better to prove a
| theorem mathematically and then see if intuition agrees. Thunder and
| lightning arrive at different times, and a child's intuition is that
| they are seperate events. An adult sees it differently. Until
| Copernicus, intuition told us the Earth is at the centre if the
| universe. After all, we see the sun cross the sky daily, it MUST be
| going around us. With greater knowledge we revise our view that we
| are turning toward and away from the sun. Never trust intuition, it
| is bane of science and the boon of religion.
|
| But aren't you using intuition to discard relativistic addition of
| velocities in your c'=c+v (or is that c=c'+v?).
|
|
| daveL

You'd really need to ask Jim that. You have my response to him (above)
confused with his statement x' = c+v.

However, I will state that the vector addition of velocities, c+v, is used
by Einstein in his equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
following which he clearly states
"But the ray moves relatively to the initial point of k,
when measured in the stationary system,
with the velocity c-v, so that x'/(c-v) = t."
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/

So you'd need to ask Einstein where he gets his "velocity c-v" from and why
he is not using (c+w)/(1+w/c) = c in that equation.

For myself, I'm wondering why the idiot couldn't make up his tiny ****ing
mind. Maybe 'w' and 'v' are too confusing for the moron.
Oops, sorry... didn't mean to offend your tin god...
Oh, wait! Yes I did.

Androcles.