GR frequency shift formula
mlut...@wanadoo.fr
What is the GR formula giving the frequency shift for the following
case:
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?
What is the result of the GR formula when M1 = M* ( M* = 1 solar
mass), R1 = 5 Km, M2 = 5 M*, R2 = 20 km, d = 50 km.
Thank you,
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1143818636.6...@t31g2000cwb.googlegroups.com...
http://groups.google.com/group/sci.physics.relativity/msg/c7984be65e7bf59b
>
> Thank you,
Always a pleasure.
Dirk Vdm
Hexenmeister
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:%5cXf.347640$Pf2.10...@phobos.telenet-ops.be...
Face it, Dork, you have a shithead, probably even stinkier
than Phuckwit Duck's.
http://www.androcles01.pwp.blueyonder.co.uk/Dork/real.htm
http://www.androcles01.pwp.blueyonder.co.uk/Dork/shit.htm
Always a pleasure to play a nasty and unfair game with an
obnoxious tord like you.
Androcles.
mlut...@wanadoo.fr
I am looking forward to a response from a GR expert, not from a mental
patient (see http://perso.wanadoo.fr/mluttgens/Vdm.htm )
Marcel Luttgens
Koobee Wublee
> case:
>
> Two non-rotating spherical celestial objects 1 and 2 are orbiting along
> their common center of gravity.
GR is the most amazing theory ever. It is the only theory out there
which can predict all the following.
** Gravitational red shift through time dilation, g_00
** Gravitational blue shift through radial wavelength, 1 / g_00
** Gravitational no shift through tangential wavelength, 1
Where
** ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2
So, it depends on the experimental data. Whatever it comes out with,
GR can predict it. On a historical note, gravitational red shift was
predicted by Einstein after he declared the Equivalence Principle. It
is nothing more than an event when Einstein finally understood gravity
= acceleration and acceleration = gravity just as Newton had said many
centuries ago. From his derivation of Doppler shift after peeking at
Voigt's paper, Einstein somehow decided v^2 is sort of the same as
acceleration. He thus decided gravitational acceleration as giving the
same effect as if the source of gravitation (the sun, the planet, etc.)
is moving away. Galilean mechanics would tell you that velocity is not
the same as acceleration. This is as basic in the study of physics as
it can get, and yet Einstein screwed it up.
While the faithful believers of GR are patting on each other's back for
the general belief, let's be a little more scientific about addressing
this issue. As you know, energy must be conserved. As Planck has
pointed out which Einstein re-iterated,
E = h f
A photon must retain its energy. Yet, Einstein's wrong analysis
actually gave the correct result from various modern experiments.
While the faithful would argue for Einstein's gifted intuition in this
case, something else must be at work here. The true cause of
gravitational red shift is still not yet discovered.
Also, according to Lorentz Transform and thus Special Relavity which is
just an interpretation to the mathematics of Lorentz Transform itself,
your scenario with identical binary stars would certainly indicate blue
shift. Both stars (their surface) have the same gravitational
potential. Thus, gravity should play no role in this. However,
claiming a blue shift would certainly not be in agreement with actual
experimentation done if possibe.
mlut...@wanadoo.fr
Thank you!
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 ?
Notice that the stars have different masses in my scenario.
Marcel Luttgens
mlut...@wanadoo.fr
Here is my approximate solution of the following problem.
Do GR experts agree?
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?
Exemple: M1 = M* ( M* = 1 solar mass = 1.989E+33 g),
R1 = 5 Km,
M2 = 5 M*,
R2 = 20 km,
d = 50 km.
M1 = 1.989E+33 'g
M2 = 1.989E+33 * 5 'g
R1 = 500000! 'cm
R2 = 2000000! 'cm
d = 5000000! 'cm
G = 6.673E-08 'Universal gravitational constant (CGS)
c = 2.998E+10 'Speed of light in cm/s
M1 = G * M1 / c ^ 2
M2 = G * M2 / c ^ 2
Nu2/Nu1 = 1 - M1*(1/R-1/(d-R2)-1/(2 * d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
= 0.9
In this case, one has a redshift!
Marcel Luttgens
Tom Roberts
> Do I have to conclude that GR cannot numerically solve this elementary
> problem where
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.
Tom Roberts tjro...@lucent.com
Koobee Wublee
news:1143896630.7...@z34g2000cwc.googlegroups.com...
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
Hexenmeister
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:DM%Yf.62409$H71....@newssvr13.news.prodigy.com...
As far as Humpty Roberts knows
"tests of strong fields are few and far between, but there are
some:
the binary pulsars, and observations of accretion disks near black
holes"
`I don't know what you mean by "observations",' Alice said.
Humpty Roberts smiled contemptuously. `Of course you don't -- till I tell
you.
I meant "there's a nice knock-down argument for you!"' <shrug>
`But "observations" doesn't mean "a nice knock-down argument",' Alice
objected.
`When I use a word,' Humpty Roberts said, in rather a scornful tone,
<shrug>,
`it means just what I choose it to mean -- neither more nor less.' <shrug>
Humpty Roberts let out a great sigh.
" <sigh>", he said.
"The nuances of English. I was discussing the usage of words and
not the concepts they represent."
-- Tom Humpty Roberts tjro...@lucent.com
news:ZDmYf.51582$2O6....@newssvr12.news.prodigy.com
"Mathematicians and physicists are human. We share the common desire to
communicate with each other easily, accurately, and concisely -- that's
why technical vocabularies were invented." said Humpty Roberts scornfully
and pretending he is human by saying "we".
Alice pondered this for moment, then asked "Was it required to fool and
mislead the 'layman'?"
"Your problem, not mine", said Humpty Roberts, then realizing his
Freudian slip, he was pretending to be human, added "(ours).
But this technical vocabulary is not secret or unfathomable, it just
takes _STUDY_. <shrug>"
And as that other shithead Andersen says,
"But the two stars of Algol have different mass, radius and density, and the
B8 is well outside of the Roche limit of the K2, while the K2 is just at the
Roche limit of the B8. That is, the K2 fills its Roche lobe completely, and
mass is transferred to the B8. So the K2 IS torn apart and there is an
accretion disk around the B8 akin to the rings of Saturn. (This accretion
disk is not stable, though. It is a transient disk; the mass transferred
from the K2 bounces off the surface of the B8 and eventually falls back to
the surface.) "
Being a B8, the surface the K2 accretion disk bounces off looks like this:
http://sohowww.nascom.nasa.gov/hotshots/2003_11_04/c2w.gif
"I'm not an astronomer" -- Humpty Roberts
You are a fuckin' stupid opinionated idiot, Roberts. <shrug>
Androcles.
mlut...@wanadoo.fr
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 tjro...@lucent.com
mlut...@wanadoo.fr
Wonderful ! I wholly agree with you, but would somebody like Steve
Carlip
also agree?
Notice that I didn't get my working formula with GR.
Marcel Luttgens
Hexenmeister
<mlut...@wanadoo.fr> wrote in message
news:1144330136....@i40g2000cwc.googlegroups.com...
|
| Tom Roberts wrote:
| > mlut...@wanadoo.fr 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?
Humpty Roberts is totally clueless.
Humpty Roberts in Wonderland:-
| Tom Roberts tjro...@lucent.com
Newsgroups: sci.physics.relativity
From: Tom Roberts <tjrobe...@lucent.com> - Find messages by this
author
Date: Sat, 17 Sep 2005 17:57:18 GMT
Local: Sat, Sep 17 2005 6:57 pm
Subject: Re: Does the 'Curvature of Spacetime' cause gravity?
"Yes, tests of strong fields are few and far between, but there are
some:
the binary pulsars, and observations of accretion disks near black
holes
`I don't know what you mean by "observations",' Alice said.
Humpty Roberts smiled contemptuously. `Of course you don't -- till I tell
you.
I meant "there's a nice knock-down argument for you!"' <shrug>
`But "observations" doesn't mean "a nice knock-down argument",' Alice
objected.
`When I use a word,' Humpty Roberts said, in rather a scornful tone,
<shrug>,
`it means just what I choose it to mean -- neither more nor less.' <shrug>
`The question is,' said Alice, `whether you can make words mean so many
different things.'
`The question is,' said Humpty Roberts, `which is to be master -- that's
all.' <shrug>
Alice was too much puzzled to say anything; so after a minute Humpty Roberts
began again. `They've a temper, some of them -- particularly verbs: they're
the proudest -- adjectives you can do anything with, but not verbs --
however,
I can manage the whole lot of them! Impenetrability! That's what I say!'
<shrug>
"And you never responded to how a 2-d surface in a flat 4-d spacetime can
have nonzero curvature, and why that shows that the curvature of such
2-d surfaces is useless in "describing" the geometry of the 4-d
manifold...." he
droned on.
"If you say that the curvature of 2-d surfaces is useless in
"describing" the geometry of the 4-d manifold....I am willing to agree
with you. But I just wanted you people to help me visualize the
intrinsic curvature of 3-d Schw. space. I was told that the Gaussian
curvature of certain 2-d surfaces will represent the intrinsic
curvature of 3-d Schw. space. When I wanted these 2-d surfaces to be
identified, Jan PB had given some interesting suggestions. But now you
say it is *useless*....." said Alice.
"_SOME_ 2-d surfaces can be useful in describing the geometry of 4-d
spacetime, in particular those spanned by a 2-d vector space of
geodesics. But you were discussing 2-d surfaces defined by coordinates,
and _those_ are useless because coordinates are completely arbitrary,
and introducing that arbitrariness destroys their usefulness" replied Humpty
Roberts.
"That means the notion of intrinsic curvature of space is either too
complex that it cannot be visualized or it is just invalid." exclaimed
Alice.
"No. But in many cases using a ball of dust particles is a better
visualization tool than 2-d surfaces.", said Humpty Roberts, teetering
on his wall.
"Mathematically it is good enough to state that in Riemannian geometry
the Riemann tensor is non-zero. Where is the necessity of associating
it with a cooked up fictitious term 'curvature of space'? " asked Alice,
thinking of the cooked up egg she had for breakfast.
"Mathematicians and physicists are human. We share the common desire to
communicate with each other easily, accurately, and concisely -- that's
why technical vocabularies were invented." said Humpty Roberts scornfully
and pretending he is human by saying "we".
Alice pondered this for moment, then asked "Was it required to fool and
mislead the 'layman'?"
"Your problem, not mine", said Humpty Roberts, then realizing his
Freudian slip, he was pretending to be human, added "(ours).
But this technical vocabulary is not secret or unfathomable, it just
takes _STUDY_. <shrug>"
Alice then went back to say "The term *curvature* basically applies to
the bending of curves and 2-d surfaces."
Ho ho, thought Humpty Roberts, "Not in differential geometry or GR.
The term "curvature" was borrowed by analogy with 2-d surfaces, and
has come to mean the Riemann curvature tensor. That is, a manifold of
_any_ dimension with nonzero Riemann tensor is said to be curved."
and he shrugged like this :- "<shrug>"
Alice asked "Why *said* to be curved when it is actually not curved?"
Humpty Roberts let out a great sigh.
" <sigh>", he said.
"The nuances of English. I was discussing the usage of words and
not the concepts they represent."
-- Tom Humpty Roberts tjro...@lucent.com
news:ZDmYf.51582$2O6....@newssvr12.news.prodigy.com
The end.
With thanks to Lewis Carroll.
The reader should take careful note here.
Humpty Roberts is not discussing the concepts words represent, he is
discussing the meaning of words. The rest of us use a dictionary.
Professor Androcles.
Tom Roberts
> Tom Roberts wrote:
>> there is no useful approximation that applies, [...]
>
> 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.
Nonsense. In a small-field situation there clearly is an appropriate
approximation, and yours might be that (I have not checked). But not for
strong fields like your problem has. <shrug>
> Let's apply it to two well known cases: [...]
You apply it to very weak-field cases. The problem you posed is not weak
field. <shrug>
Note, for instance, that in your problem the frequency shift is not
constant (due to the emission of gravitational radiation and the
consequent inspiral of the two objects), ignoring the fact that the
observers will be occluded by the objects as they orbit.
Tom Roberts tjro...@lucent.com
mlut...@wanadoo.fr
Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Tom Roberts wrote:
> >> there is no useful approximation that applies, [...]
> >
> > 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.
>
> Nonsense. In a small-field situation there clearly is an appropriate
> approximation, and yours might be that (I have not checked). But not for
> strong fields like your problem has. <shrug>
>
On one hand, you claimed that no useful approximation apply, but on the
other hand, you now recognize that an approximate approximation exists
in small-field situations.
Remember that you added that "this is a quite complicated computation
that can only be performed via numerical simulation".
What is then your GR approximation? You are not very curious. I am
pretty well convinced that you will find back my formula. Notice that I
spoke of an *approximate* formula. You should consider the solution I
gave as a solution to a "thought experiment".
It would be nice if you could show the steps leading to your
small-field GR solution.
Marcel Luttgens
va...@cox.net
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
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.
va...@cox.net
No doubt Dirk knows how to derive the weak field approximation. He
certainly wouldn't forget to account for the distance between the
emitter and the receiver in a gravitational field [weak or strong].
>
> Marcel Luttgens
mlut...@wanadoo.fr
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.
Marcel Luttgens
va...@cox.net
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.
Using your formula the GPS would have crashed and burned in a couple of
minutes.
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"?
mlut...@wanadoo.fr
Almost right!
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.
Notice that my reduced formula is identical to the GR approximate
formula.
>
> Using your formula the GPS would have crashed and burned in a couple of
> minutes.
Seemingly, you have a problem with simplifying a formula.
> 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 ?",
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. 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.
Marcel Luttgens
va...@cox.net
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.
> > 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.
>
> 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.
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..
>
> Marcel Luttgens
Sue...
va...@cox.net wrote:
snip
> > > 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
What does GR predict for a cesium clock located in
hollow in the center of the planet ?
Sue...
va...@cox.net
Correction: ~615.4 meters
>
> 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.
Correction: Your formula is garbage. If it was identical to the GR
formula it would'nt
Hexenmeister
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1144615332.7...@v46g2000cwv.googlegroups.com...
|
| va...@cox.net wrote:
| snip
| > > > 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
|
| What does GR predict for a cesium clock located in
| hollow in the center of the planet ?
The planet doesn't have a hollow centre.
However, the idiot's imagination is running wild as well,
the true figure for the error is about half an inch, not half a mile.
http://www.androcles01.pwp.blueyonder.co.uk/GPS/sundials.htm
Androcles.
va...@cox.net
I took the time to get your math right. Your formula predicts the delta
to be
.000031622400 seconds/day
The formula derived from the Schwarzschild solution predicts the delta
to be
.000038407392 seconds/day
Your formula is incorrect by a whopping 7000 nanoseconds a day
Using your formula GPS accuracy would be off 1538.5 meters after being
in operation for 1 day.
va...@cox.net
Hexenmeister wrote:
> "Sue..." <suzyse...@yahoo.com.au> wrote in message news:1144615332.7...@v46g2000cwv.googlegroups.com...
> |
> | va...@cox.net wrote:
> | snip
> | > > > 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
> |
> | What does GR predict for a cesium clock located in
> | hollow in the center of the planet ?
>
> The planet doesn't have a hollow centre.
> However, the idiot's imagination is running wild as well,
> the true figure for the error is about half an inch, not half a mile.
Marcels delta period is incorrect by 7.853E-11. The error becomes
.023542701 m/s. Ater 86400 s the error becomes 2034 m. The GPS as a
functioning tool would have failed early in the first day of use. After
15 minutes the error would have been 21.2 m.
Androcles: Stick to acting like a jerkoff since that seems to be all
you do well.
mlut...@wanadoo.fr
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
mlut...@wanadoo.fr
Thank you! I already apologized for having taken 20200 km as the
orbital radius of GPS satellites, whereas it is their altitude. Iow,
the radius is 20200 km + the Earth radius.
In my formula, d represents the orbital radius, not the altitude.
In my formula, the distance between centers is d,
meaning that the orbital radius of the GPS satellite whith altitude
20200 km is d = 20200 km + the Earth radius = ~26578 km.
Remember that my general formula is:
Nu2/Nu1 = 1 - M1*(1/R-1/(d-R2)-1/(2 * d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
where, in this simple case of GPS satellites,
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 + Earth radius 'Orbital radius of the satellite in cm
= ~2,658E+9 cm
For GPS satellites, my formula reduces to
Nu2/Nu1 = 1 + M2*(1/R2-1/d-1/2d), and gives a
shift = Nu2/Nu1 - 1 = 4.45E-10
Multiplying that shift with the day in microseconds, one gets 38.5
microseconds/day,
the same value as the one obtained with the approximate GR formula.
Notice that my "general" formula is approximately valid for all fields,
thus also the so-called strong ones.
Try the GR numerical simulation to solve the more general problem where
M1 = M* ( M* = 1 solar mass), R1 = 5 Km, M2 = 5 M*, R2 = 20 km, d = 50
km, before
calling my formula garbage.
You should find Nu2/Nu1 = ~0.9
Marcel Luttgens
va...@cox.net
You finally figured out your mistake. After telling me the GR formula
was wrong several times. Good for you. I suspect you would still think
your original formula was correct if I hadn't shown you the simple
approximation derived from the Schwarzschild metric and informed you of
the actual value of the correction used for the GPS. Apparently you
just assumed that the actual correction must correspond with your
initial numerical value.
Luttgen's pudding is made of puka. His judgement of GR's value, as a
scientific tool, is irrelevent. He doesn't understand how the theory
can be used, since he hasn't opened the instruction manual. If he knew
how the theory could be used he wouldn't have asked the original
questions. If it wasn't for the introduction of GR he would still be
claiming the delta period correction for the GPS was 31 thousand
nanoseconds per day. I looked at luttgen's homepage and it seems his
entire purpose is to make ignorant comments about relativistic physics.
Bruce
Tom Roberts
> Tom Roberts wrote:
>>In a small-field situation there clearly is an appropriate
>>approximation, and yours might be that (I have not checked). But not for
>>strong fields like your problem has. <shrug>
>
> On one hand, you claimed that no useful approximation apply, but on the
> other hand, you now recognize that an approximate approximation exists
> in small-field situations.
Either you are not paying attention, or you are attempting to create
controversy where none exists. I have always known that a small-field
approximation exists. It's just that your original problem is not
small-field, and no useful approximation exists for it. <shrug>
> What is then your GR approximation?
Look above -- there is no useful GR approximation for this problem. <shrug>
> It would be nice if you could show the steps leading to your
> small-field GR solution.
The small-field GR approximation is not "mine", and is presented in
every GR textbook. Study.
Tom Roberts tjro...@lucent.com
mlut...@wanadoo.fr
Sue... wrote:
> va...@cox.net wrote:
> snip
> > > > 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
>
> What does GR predict for a cesium clock located in
> hollow in the center of the planet ?
>
> Sue...
I have no idea of what GR would predict.
As for me, the shift is -3.47E-10
Marcel Luttgens
Sue...
mlut...@wanadoo.fr wrote:
> Sue... wrote:
> > va...@cox.net wrote:
> > snip
> > > > > 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
> >
<< What does GR predict for a cesium clock located in
hollow in the center of the planet ? >>
> >
> > Sue...
>
>
<<
> 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. >>
> I have no idea of what GR would predict.
> As for me, the shift is -3.47E-10
You seem to be saying it would run slower than a
a clock at the surface at a point with less gravity
than at the surface. Hmmm ?
Sue...
mlut...@wanadoo.fr
Sue
Marcel Luttgens
You seem to be saying it would run slower than a
a clock at the surface at a point with less gravity
than at the surface. Hmmm ?
Sue...
The shift is of course 3.47E-10 !
Marcel Luttgens
> >
> > >
> > > >
> > > > > 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.
> > > >
mlut...@wanadoo.fr
"What does GR predict for a cesium clock located in
hollow in the center of the planet ?"
Luttgens:
I have no idea of what GR would predict.
As for me, the shift is -3.47E-10
Sue:
You seem to be saying it would run slower than a
a clock at the surface at a point with less gravity
than at the surface. Hmmm ?
Luttgens:
The shift is of course 3.47E-10 !
In fact, if a light of frequency Nu1 is emitted from the center of the
Earth, the corresponding frequency Nu2 of the light received at the
surface (through a hole) is given by the relation Nu2/Nu1 = 1 -
GM/2Rc^2, where M and R are respectively the mass and the radius of the
Earth.
The observed redshift is thus -GM/2Rc^2 = -3.47E-10.
This would imply that a clock at the center is *observed* to run slower
than a clock at the surface, even if the gravitational potential is
zero at the center.
Marcel Luttgens
Sue...
mlut...@wanadoo.fr wrote:
> Sue:
>
> "What does GR predict for a cesium clock located in
> hollow in the center of the planet ?"
>
> Luttgens:
>
> I have no idea of what GR would predict.
> As for me, the shift is -3.47E-10
>
> Sue:
>
> You seem to be saying it would run slower than a
> a clock at the surface at a point with less gravity
> than at the surface. Hmmm ?
>
> Luttgens:
>
> The shift is of course 3.47E-10 !
>
> In fact, if a light of frequency Nu1 is emitted from the center of the
> Earth, the corresponding frequency Nu2 of the light received at the
> surface (through a hole) is given by the relation Nu2/Nu1 = 1 -
That is disproven by Pound/Snider and Mosbauer spectrometry.
> GM/2Rc^2, where M and R are respectively the mass and the radius of the
> Earth.
> The observed redshift is thus -GM/2Rc^2 = -3.47E-10.
> This would imply that a clock at the center is *observed* to run slower
> than a clock at the surface, even if the gravitational potential is
> zero at the center.
Review the doppler mechanisms used Mossbauer spectrometry.
No asterisks are necessary using the word *observed*.
The shifts in nuclear resonance are accurately measured.
Sue...
>
> Marcel Luttgens
mlut...@wanadoo.fr
What I meaned is that a real shift is observed, but this doesn't mean
that the clock at the center of the Earth ticks really slower that a
clock at the surface.
More generally, the frequency shift of the light emitted by a satellite
or from the Earth center is *interpreted* as a clock slowing due to the
local gravitational potential (for instance, zero at the Earth center).
In fact, the observed shift is easily explained by a gravitational gain
or loss of energy of the emitted light during its journey to the
receiver.
Marcel Luttgens
> Sue...
>
> >
> > Marcel Luttgens
Sue...
Why not simply say the nuclear resonance is higher or lower
instead insted of *interpreting* causality violations?
An EM path can't create clocks ticks that didn't occur
(static blue shift) nor can it dispose of ticks that did
occur. (static redshift).
"On the Interpretation of the Redshift in a Static Gravitational Field"
http://arxiv.org/abs/physics/9907017
http://www.ds.mw.tu-darmstadt.de/magnetism/content/mbeffect.html
Sue...
mlut...@wanadoo.fr
There are two interpretations:
1) Clocks 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. The light thus appears to be redshifted relative to
the frequency of the clock.
This is the GR interpretation.
2) On the other hand the phenomenon is alternatively discussed (even in
some authoritative texts) in terms of an energy loss of a photon as it
overcomes the
gravitational attraction of the massive body.
This is also my interpretation, because a photon has some energy E =
hNu, to which corresponds a pseudo-mass hNu/c^2.
Then, it is possible to calculate the energy loss of a photon coming
from the center of the Earth, leading to a redshift of GM/2Rc^2 =
-3.47E-10.
>From such redshift, one could infer that a clock situated at the center
of the Earth run slower than a clock at the surface, whereas, according
to GR, it should run faster, because the gravitational potential at the
center is less than at the surface.
Btw, would you agree, using GR, that the redshift is -3.47E-10?
> http://www.ds.mw.tu-darmstadt.de/magnetism/content/mbeffect.html
>From that paper, I retain:
"That is, it was found that a photon of energy E behaves as though it
had a mass of E/c2, in which c is the velocity of light."
I have to conclude that the GR interpretation, according to which "the
energy and frequency of the propagating photon do not change with
height. The light thus appears to be redshifted relative to the
frequency of the clock." is wrong .
Marcel Luttgens
>
> Sue...
>
>
> >
> > Marcel Luttgens
> >
> > > Sue...
> > >
> > > >
> > > > Marcel Luttgens
Sue...
But you don't know a photon's path or probability of absorbtion.
>
> Btw, would you agree, using GR, that the redshift is -3.47E-10?
I would agree there is no frequency shift between co-moving
EM coupling structures.
>
> > http://www.ds.mw.tu-darmstadt.de/magnetism/content/mbeffect.html
>
> >From that paper, I retain:
>
> "That is, it was found that a photon of energy E behaves as though it
> had a mass of E/c2, in which c is the velocity of light."
>
> I have to conclude that the GR interpretation, according to which "the
> energy and frequency of the propagating photon do not change with
> height. The light thus appears to be redshifted relative to the
> frequency of the clock." is wrong .
Pound Snider used the term nuclear resonance in the later
paper to avoid that confusion.
Sue...
>
> Marcel Luttgens
>
>
> >
> > Sue...
> >
> >
> > >
> > > Marcel Luttgens
> > >
> > > > Sue...
> > > >
> > > > >
> > > > > Marcel Luttgens
mlut...@wanadoo.fr
Which path? Absorbtion by what?
>
> >
> > Btw, would you agree, using GR, that the redshift is -3.47E-10?
> I would agree there is no frequency shift between co-moving
> EM coupling structures.
I don't understand. Would you observe another redshift? Which one?
>
> >
> > > http://www.ds.mw.tu-darmstadt.de/magnetism/content/mbeffect.html
> >
> > >From that paper, I retain:
> >
> > "That is, it was found that a photon of energy E behaves as though it
> > had a mass of E/c2, in which c is the velocity of light."
> >
> > I have to conclude that the GR interpretation, according to which "the
> > energy and frequency of the propagating photon do not change with
> > height. The light thus appears to be redshifted relative to the
> > frequency of the clock." is wrong .
>
> Pound Snider used the term nuclear resonance in the later
> paper to avoid that confusion.
Which confusion? That a photon of energy E doesn't behave as if it had
a mass of E/c2?
Marcel Luttgens
>
> Sue...
Sue...
Photons only have definition at the instant of emission
and absorbtion by atomic oscillators. They are not valid
propagation models. That is why Feynman equips them
with wrist watches and magnetic monopoles to make QED work.
And told sudents to "shut up and calculate" when they
asked which arm the watch was worn on. :o)
http://www.physics.yorku.ca/undergrad_programme/highsch/Feynm4.html
http://nobelprize.org/physics/articles/ekspong/index.html
http://nobelprize.org/physics/laureates/1965/feynman-lecture.html
>
> >
> > >
> > > Btw, would you agree, using GR, that the redshift is -3.47E-10?
>
> > I would agree there is no frequency shift between co-moving
> > EM coupling structures.
>
> I don't understand. Would you observe another redshift? Which one?
Redshift is a misleading term as the paper:
http://arxiv.org/abs/physics/9907017 explains
It is normally applied for relative motion as
in the Hubble law. Sometimes confused with
reddening which involves atomic absorbtion and
reemission by interstellar particles.
>
> >
> > >
> > > > http://www.ds.mw.tu-darmstadt.de/magnetism/content/mbeffect.html
> > >
> > > >From that paper, I retain:
> > >
> > > "That is, it was found that a photon of energy E behaves as though it
> > > had a mass of E/c2, in which c is the velocity of light."
> > >
> > > I have to conclude that the GR interpretation, according to which "the
> > > energy and frequency of the propagating photon do not change with
> > > height. The light thus appears to be redshifted relative to the
> > > frequency of the clock." is wrong .
> >
> > Pound Snider used the term nuclear resonance in the later
> > paper to avoid that confusion.
>
> Which confusion? That a photon of energy E doesn't behave as if it had
> a mass of E/c2?
At the instant of emission or absorbtion the energy is defined by
the atomic oscillator... not by the incident light. The absorbed
energy may not even be from an atom but rather a microwave tube
which operates without any atomic emission or some other
spurious source.
So when the absorbing atom is near the earth, its lower
frequency light has to be Doppler shifted up, to match a
similar atom at higher altitude and be absorbed.
The nuclear resonance and hyperfine transitions of atoms
are affected by gravity but is mathematically absurd to
attribute that to a light path. It would be a violation of
causality.
Sue...
>
> Marcel Luttgens
>
> >
> > Sue...
Daryl McCullough
>This is also my interpretation, because a photon has some energy E =
>hNu, to which corresponds a pseudo-mass hNu/c^2.
>Then, it is possible to calculate the energy loss of a photon coming
>from the center of the Earth, leading to a redshift of GM/2Rc^2 =
>-3.47E-10.
Where is the 2 coming from? It should be GM/Rc^2 = 7E-10.
>From such redshift, one could infer that a clock situated at the center
>of the Earth run slower than a clock at the surface, whereas, according
>to GR, it should run faster, because the gravitational potential at the
>center is less than at the surface.
GR predicts that the clock at the surface runs faster than
the clock at the center. The clock with the *greater* potential
runs faster.
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
John C. Polasek
McCullough) wrote:
A clock at infinity is fastest, losing 7e-10 by the time it is taken
down to the surface of the earth. Taking it down to the center it
loses another 7e-10 or 1.4e-9 total. The clock does.
Its wavelength at the clock stays the same because c reduces in the
same proportion. But the radiation once emitted from heavy gravity to
lighter has its wavelength stretched on the way so you would see a red
shift or longer wavelength, but its frequency remains the same.
(The gravity inside the earth is proportional to radius: g = g0*r/R
and the work done is g0*R^2/2R).
John Polasek
http://www.dualspace.net
Sue...
Tom Roberts
> There are two interpretations:
> 1) Clocks 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. The light thus appears to be redshifted relative
> to the frequency of the clock.
>
> This is the GR interpretation.
Not really. That is merely one possible interpretation in GR.
> 2) On the other hand the phenomenon is alternatively discussed (even
> in
> some authoritative texts) in terms of an energy loss of a photon as it
> overcomes the gravitational attraction of the massive body.
That is another possible interpretation in GR.
The point is that GR itself is independent of coordinate choice, but
clock rate, energy, and frequency are not. These two interpretations are
based on different choices of coordinates, that's all.
[There are also possible interpretations in which both the
clock rates vary and the energy/frequency of the light vary.]
Note in either interpretation the actual _measurements_ remain the same.
That's because all physical measurements yield values that are
independent of coordinates.
Tom Roberts tjro...@lucent.com
Sue...
T E Cranshaw1 and J P Schiffer2
1 U.K. Atomic Energy Research Establishment, Harwell, Berks.
2 Argonne National Laboratory, Argonne, Illinois, U.S.A., and U.K.
Atomic Energy Research Establishment, Harwell, Berks.
Abstract. The shift in frequency of electromagnetic radiation as it
passes through a difference in gravitational potential has been
measured by use of the Mössbauer effect in 57Fe. The result is 0.859
± 0.085 times the value predicted by Einstein's theory of relativity,
the error being a standard deviation. The discrepancy is most likely
due to chance since the probability of obtaining a deviation of at
least this magnitude is 10%.
Print publication: Issue 2 (August 1964)
Received 4 February 1964
http://www.iop.org/EJ/abstract/0370-1328/84/2/307/
Hmmmm...
Sue...
John C. Polasek
wrote:
Why don't you put some numbers to your theories? The thread originated
that way. Your jousting is wearying and, frankly, convey little.
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay
mlut...@wanadoo.fr
Iow, GR cannot be proven or disproved by the observed shift.
By the way, if a light of frequency Nu1 is emitted from the center of
the Earth, what would be the formula giving the light frequency Nu2
observed at the surface of the Earth?
And what would be the observed shift?
Daryl McCullough
>
>mlut...@wanadoo.fr says...
>
>>This is also my interpretation, because a photon has some energy E =
>>hNu, to which corresponds a pseudo-mass hNu/c^2.
>>Then, it is possible to calculate the energy loss of a photon coming
>>from the center of the Earth, leading to a redshift of GM/2Rc^2 =
>>-3.47E-10.
>
>Where is the 2 coming from? It should be GM/Rc^2 = 7E-10.
Actually, I'm wrong, and the 2 is right.
The potential difference between the center
of the Earth and the surface is
- integral of GM/R^3 r dr
= -1/2 GM/R
John C. Polasek
You have to keep your eye on the clock as well as the light. Take a
clock from infinity to earth and it loses frequency by 7e-10. Take it
down to the center and it loses another 3.5e-10 (sorry I dropped a 2
before).
So the clock is slow by 3.5e-10 at center and thus its frequency,
relative to earths surface. Nu1 remains low on its way to the surface,
neither gaining nor losing energy E = hnu so Nu2 = Nu1.
But light speeds up out of gravity, so it will suffer a 3.5e-10
redshiftstretch in wavelength at the surface. At infinity it will have
stretched 10.5e-10.
Relativity treats this as time dilation with constant c, and without
going through the motions, you wind up with a paradox if light loses
energy on the way up.
>Marcel Luttgens
>
>>
>> Tom Roberts tjro...@lucent.com
John Polasek
Pmb
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:H6C%f.65501$H71....@newssvr13.news.prodigy.com...
> mlut...@wanadoo.fr wrote:
> > There are two interpretations:
> > 1) Clocks 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. The light thus appears to be redshifted relative
> > to the frequency of the clock.
> >
> > This is the GR interpretation.
>
> Not really. That is merely one possible interpretation in GR.
>
>
> > 2) On the other hand the phenomenon is alternatively discussed (even
> > in
> > some authoritative texts) in terms of an energy loss of a photon as it
> > overcomes the gravitational attraction of the massive body.
>
> That is another possible interpretation in GR.
Please give another one.
> The point is that GR itself is independent of coordinate choice, ...
When one wants to atually use GR, i.e. take a measurement, then one must
resort to components (i.e. scalar product with observer's 4-velocity etc)
> ...but clock rate, energy, and frequency are not. These two
> interpretations are based on different choices of coordinates, that's all.
That is not another interpretation. mlut...@wanadoo.fr was pretty clear
as to what he meant and that is in keeping with GR. In fact this is how
Einstein first obtained gravitational redshift - He choose two coordinate
system and evalualted the frequency in each one.
Pete
mlut...@wanadoo.fr
I am looking forward to the explanation and formula, that could be
given by
Tom Roberts.
Imo, the clock keeps ticking at the same rate, whether it is situated
at infinity
or at the center of the Earth.
> So the clock is slow by 3.5e-10 at center and thus its frequency,
> relative to earths surface. Nu1 remains low on its way to the surface,
> neither gaining nor losing energy E = hnu so Nu2 = Nu1.
> But light speeds up out of gravity, so it will suffer a 3.5e-10
> redshiftstretch in wavelength at the surface. At infinity it will have
> stretched 10.5e-10.
> Relativity treats this as time dilation with constant c, and without
> going through the motions, you wind up with a paradox if light loses
> energy on the way up.
A paradox for those who assume that clocks at rest are affected by
gravity.
Marcel Luttgens
va...@cox.net
The formula can be reviewed at this site. Chapter 9, problem 9.2.5.
http://www.geocities.com/zcphysicsms/chap9.htm
Note at the center spacetime curvature vanishes but gravitational time
dilation doesn't.
Converting to geometric units and approximating
dTau = [ 1 + Mr^2/R^3 - 3M/R ]^1/2 dt
to
dTau = [ 1+ (1/2)(Mr^2/R^3) - 3M/R ] dt
setting r ~ R
dTau = [ 1 + (M/2R) - (3M/2R) ] dt
The calculation reveals that a clock at the center of the Earth will
actually record less ticks than a clock on the surface of the Earth.
The delta period is ~ 6.82E-17. Over 4.5 billion years this amounts to
9.68 seconds. If you read this interesting pedagogical paper Okun and
Selivanov mention something Feynman said during his lectures on
gravitation. Last paragraph of page 3.
http://arxiv.org/PS_cache/physics/pdf/9907/9907017.pdf
> Imo, the clock keeps ticking at the same rate, whether it is situated
> at infinity
> or at the center of the Earth.
Since the Newtonian potential at the Earth's center is the same as the
boundary condition you might think so. This is another reason why we
need to use GR to make the correct prediction.
Bruce
va...@cox.net
Should read
dTau = [ 1+ (1/2)(Mr^2/R^3) - (3/2)(M/R) ] dt
John C. Polasek
snip
Polasek says:
Your reference to Ch. 9 pathetically demonstrates of the futility of
GR. Is that chapter supposed to be of tutorial value or were you just
showing how smart you are?
Let me show you how easy it is to figure out df/f in Dual Space
relativity. (= 3.4e-10).
Visit my website paper #2, "Dual Space gravity" and see that I have
added a couple of terms to Newton's law. The term cdc/dr = g shows how
the velocity of light varies in a g-field. The frequency does exactly
the same. Follow the arithmetic:
Gravity inside the earth goes as
gin = g0*r/R = MGr/R^3
From Eq. 1 of my gravity paper we have
cdc/dr = gin
Rearrange to find the fractional change from center to surface:
dc/c = gin*dr/c^2 = MGrdr/R^3c^2
Integrate from 0 to R to get dc/c = 3.4754 x 10^-10, a figure
previously mentioned from relativity, the hard way.
The fractional gain in c equals the fractional gain in nu.
Eq. 1 shows the cause of gravity, namely the stealing of material at
creation time causing weakening of "Espace", the creator's stockpile.
For this loss of "vigor" both c and nu are reduced, according to the
Navier Stokes equation on which Eq. 1 is based.
Dual Space relativity is vastly better than your brand, and it makes
sense, and it works in Euclidean space. Incidentally, it is also the
basis for my new solution to the Pioneer Anomaly which has so far not
been solved by anyone.
Thank you for your kind attention.
John Polasek
http://www.dualspace.net
va...@cox.net
empirically roundfiled. Eq.1 shows just how clueless a crank you are.
mlut...@wanadoo.fr
There seems to be a logical bug in GR.
Let's consider the following experiment:
1) A light ray of frequency NuO is sent from a point O situated at
the Earth surface to a point A situated at the summit of a tower
whose height h = 100 m.
At O, the gravitational potential is of course gO = GM/R^2,
where G is the gravitational constant, M the mass of the Earth and
R the Earth radius.
At A, the gravitational potential is gA = GM/(R+h)^2.
As gA < gO, a clock at A ticks faster than a clock at O (according
to GR).
As gA is very close to gO, the frequency NuA of the light observed
at A is about NuA / NuO = 1 - gO h/c^2. Thus, at A, a redshift
is observed, and the A observer will conclude that its clock ticks
faster than a surface clock.
2) The same light is sent from O to the bottom B of a pit whose
depth d = 100 m.
At B, the gravitational potential is gB = GM/R^2 * (R-d)/R.
As gB < gO, a clock at B ticks faster than a clock at O (according
to GR).
As gB is very close to gO, the frequency NuB of the light observed
at B is about NuB / NuO = 1 + gO d/c^2. Thus, at B, a blue shift
is observed, and the B observer will conclude that its clock ticks
slower than a suface clock, which contradicts the GR claim that
it should tick faster.
The only logical conclusion is that clocks O, A and B tick
intrinsically
at the same rate (contrarily to the GR claim), but due to the energy
gain or loss of the photon during its travel to A or B, the surface
clock will appear to tick at different rates according to the situation
of the observer.
Marcel Luttgens
mlut...@wanadoo.fr
I presume that r is the distance of the clock from the center of the
Earth.
Am I right?
Marcel Luttgens
John C. Polasek
>If your gravity paper makes a prediction different than GR then it is
>empirically roundfiled. Eq.1 shows just how clueless a crank you are.
First I was going to respond that you apparently didn't know whether
my answer was correct or not, and then I saw that you went through
some GR incanctations to get a meaningless result:
"The delta period is ~ 6.82E-17. Over 4.5 billion years this amounts
to 9.68 seconds".
Is that what your GR gives you? What does it mean? I already showed
you the fractional increase in frequency from center to surface is
3.4e-10, and earlier someone had the same result.
Did you look at Eq. 1 and the explanation for it? It gives a logical
reason for gravity. You don't really think the void of space can be
bent, do you? God can't do it. Only Minkowski can.
John Polasek
http://www.dualspace.net
Tom Roberts
>> same.
>> That's because all physical measurements yield values that are
>> independent of coordinates.
>
> Iow, GR cannot be proven or disproved by the observed shift.
No physical theory can be "proven or disproved" in any way -- those
concepts are from mathematics and logic, and simply do not apply to
physical theories.
GR could be _refuted_ by such measurements, because it makes a definite
prediction for the _MEASUREMENT_. So far, no such measurement like this
has done so, and a number of them have confirmed the GR prediction.
Do not confuse interpretations with measurements. Interpretations are
fungible; measurements are definite. There may be many possible
interpretations of a theory, but only a single prediction for a given
measurement.
> By the way, if a light of frequency Nu1 is emitted from the center of
> the Earth, what would be the formula giving the light frequency Nu2
> observed at the surface of the Earth?
To excellent approximation, compute the Newtonian gravitational
potential at the source and the detector. Then:
f_det = sqrt((1+2P_detector/c2)/(1+2P_source/c2)) f_source
where f_x is the frequency at x and P_x is the Newtonian gravitational
potential for x (remember it is negative). This neglects the rotation of
the earth, and assumes both source and detector are at rest relative to
the rotationless earth.
For more general situations one must first obtain the metric, and then
integrate it for the conditions of the measurement.
> And what would be the observed shift?
The actual measurement is impossible to do.
Tom Roberts tjro...@lucent.com
Tom Roberts
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:H6C%f.65501$H71....@newssvr13.news.prodigy.com...
>
> When one wants to atually use GR, i.e. take a measurement, then one must
> resort to components (i.e. scalar product with observer's 4-velocity etc)
Sure. So what? -- the _theory_ is independent of your choice.
>> ...but clock rate, energy, and frequency are not. These two
>> interpretations are based on different choices of coordinates, that's all.
>
> That is not another interpretation. mlut...@wanadoo.fr was pretty clear
> as to what he meant and that is in keeping with GR.
He was wrong in claiming that one and only one interpretation was "_THE_
GR interpretation [emphasis mine]".
Sure, what he said is "in keeping with GR". But so is the other
interpretation he mentioned, as are a zillion other potential ones.
> In fact this is how
> Einstein first obtained gravitational redshift - He choose two coordinate
> system and evalualted the frequency in each one.
Right -- he chose coordinate systems and did the calculations using
them. Had he chosen other coordinate systems the details of the algebra
would have been different, and the interpretation of the method would
have been different, but the prediction for the measurement would have
been the same.
This computation is easy to describe in coordinate-independent terms,
showing that the dependence on coordinates is not intrinsic:
Take the 4-velocity of the source, parallel propagate it
along the null geodesic followed by the light to the
detector, and take the dot product with the 4-velocity
of the detector. The result is a real number representing
the ratio of the source-measured frequency to the
detector-measured frequency, where both measurements are
made using a standard clock collocated with the measurement.
Tom Roberts tjro...@lucent.com
John C. Polasek
wrote:
>Pmb wrote:
Well, I see nobody has the answer yet for the fractional change in c
from the center of the earth to its surface.
So for extra credit, part B, calculate the fractional change from the
surface to infinity using of course GTR.. Just to help out, I'll show
you how I do it using my modified Newton equation #1:
Follow the arithmetic:
Gravity outside the earth goes as
gout(r) = MG/r^2
From Eq. 1 of my gravity paper we have
cdc/dr = gout(r)
Rearrange to find the fractional change from surface to infinity:
dc/c = gout *dr/c^2 = MG/r^2c^2
Integrate from 0 to R to get dc/c = 6.950 x 10^-10, twice the inside
value.
The velocity of light increases by 3.5e-10 + 7.0e-10 from inside the
earth to infinity. The frequency does not change. The wavelength
stretches.
The ambivalence shown above, that either of the interpretations might
be correct does no credit to relativity. Pound and Rebka still have
the "Weight of the Photons" for one view, and the Shapiro effect as a
decrease of velocity for another. Most think the frequency is worn
down by gh/c^2.
And therein is the paradox. If the frequency loses by fatigue on the
way up, what are we to say when the comparison clock on top is now
running faster by the same fraction? We have to accept that there's a
double red shift. This is not discussed in polite company.
John Polasek
http://www.dualspace.net
va...@cox.net
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. Read: Okun and Selivanov
pedagogical paper entitled "On the interpretation of the redshift in a
static gravitational field".
http://arxiv.org/PS_cache/physics/pdf/9907/9907017.pdf
Bruce
Tom Roberts
> 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.
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 tjro...@lucent.com
va...@cox.net
Tom Roberts wrote:
> va...@cox.net 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.
mlut...@wanadoo.fr
You wrote: "The actual measurement is impossible to do."
Yes, but you could nevertheless use your formula and get a theoretical
answer.
Anyhow, your formula
f_det = sqrt((1+2P_detector/c2)/(1+2P_source/c2)) f_source
doesn't apply to the Earth interior. It leads to a result which is
wrong by a
factor 2.
mlut...@wanadoo.fr
va...@cox.net wrote:
> Tom Roberts wrote:
> > va...@cox.net 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."
Moreover, GR is simply not applicable to the Earth interior.
Marcel Luttgens
va...@cox.net
mlut...@wanadoo.fr wrote:
> va...@cox.net wrote:
> > Tom Roberts wrote:
> > > va...@cox.net 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
Pmb
<va...@cox.net> wrote in message
news:1145219736.5...@g10g2000cwb.googlegroups.com...
>
> mlut...@wanadoo.fr wrote:
>> va...@cox.net wrote:
>> > Tom Roberts wrote:
>> > > va...@cox.net 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.
That is a bit off. As measured from a *single* coordinate system all clocks
run at the same rate. E.g. All clocks tick off at the same rate of
*cooprdinate time* but different rates of *proper time*. If I change my
coordinate system such that the physical origin (where the observer is
located) then the clock at his location ticks off coordinate time according
to the rate that the clock at the origins clock ticks off proper time.
For those interested see "On the Interpretation of the Redshift in a Static
Gravitational Field" by Lev Okun et.al. at
http://xxx.lanl.gov/abs/physics/9907017
> 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
Bruce. How do you like your new test by Schutz?
Pete
John C. Polasek
>
><va...@cox.net> wrote in message
>news:1145219736.5...@g10g2000cwb.googlegroups.com...
>>
>> mlut...@wanadoo.fr wrote:
>>> va...@cox.net wrote:
>>> > Tom Roberts wrote:
>>> > > va...@cox.net 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.
>
>That is a bit off. As measured from a *single* coordinate system all clocks
>run at the same rate. E.g. All clocks tick off at the same rate of
>*cooprdinate time* but different rates of *proper time*. If I change my
>coordinate system such that the physical origin (where the observer is
>located) then the clock at his location ticks off coordinate time according
>to the rate that the clock at the origins clock ticks off proper time.
>
>For those interested see "On the Interpretation of the Redshift in a Static
>Gravitational Field" by Lev Okun et.al. at
>http://xxx.lanl.gov/abs/physics/9907017
>
The Okun article is carefully written and suitably abstruse as GR
deserves, and recognizes that you will get a double redshift with
Pounds' heavy photon.
But there is one little hitch. He correctly sticks with a slow low
clock, a fast up clock, no change in frequency, the radiated frequency
comparing low with the up clock, but of course with c remaining
constant.
The rub is that there's no redshift since both c and f are unchanged:
Lam = c/f
There is no proper lengthening of the wavelength. Redshift is the
movement of characteristic lines toward the longer wave. It seems like
it doesn't happen here.
(What's wrong of course is sticking to time dilation instead of
noticing that c itself is reduced by gravity). dc/dr = MG/r^2c (Eq.
1).
snip
mlut...@wanadoo.fr
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.
Marcel Luttgens
Daryl McCullough
>That is a bit off. As measured from a *single* coordinate system all clocks
>run at the same rate. E.g. All clocks tick off at the same rate of
>*cooprdinate time* but different rates of *proper time*.
I'm not exactly sure what you are saying in that paragraph, but
it seems backwards to me. If you have an accurate clock that "ticks"
once per second, then moving it from sea level to the top of a tall
mountain has no effect on the *proper* time between ticks. The
proper time will still be 1 second between clicks. However, the
*coordinate* time between clicks will vary depending on the altitude
of the clock.
va...@cox.net
> <va...@cox.net> wrote in message
> news:1145219736.5...@g10g2000cwb.googlegroups.com...
> >
> > mlut...@wanadoo.fr wrote:
> >> va...@cox.net wrote:
> >> > Tom Roberts wrote:
> >> > > va...@cox.net 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.
>
> That is a bit off. As measured from a *single* coordinate system all clocks
> run at the same rate. E.g. All clocks tick off at the same rate of
> *cooprdinate time* but different rates of *proper time*. If I change my
> coordinate system such that the physical origin (where the observer is
> located) then the clock at his location ticks off coordinate time according
> to the rate that the clock at the origins clock ticks off proper time.
I did a poor job of stating the case. What I meant is
dt_shell = ( 1 - 2M/r)^1/2 dt
dt = bookkeeper coordinate time reckoned for the GR boundary condition.
ie flat spacetime at infinity [ far awy ].
dt_shell = proper time measured at r_shell.
To find a ratio for dt_shell 1 and 2
dt_shell 1 / ( 1 - 2M_1/r_1 )^1/2 = dt = dt_shell 2 / ( 1 - 2M_2/r_2
)^1/2
dt_shell 2 / dt_shell 1 = ( 1 - 2M_2/r_2 )^1/2 / ( 1 - 2M_1/r_1 )^1/2
For this coordinate system all clocks located on a specific r_shell
will tick at the same rate. As you stated. Clocks located on different
r_shell will tick at different rates. Thats what I meant when I said
clocks located at different positions in the gravitational field will
tick at different rates. I should have said potential rather than
position since all the clocks could be at the same r_shell.
>
> For those interested see "On the Interpretation of the Redshift in a Static
> Gravitational Field" by Lev Okun et.al. at
> http://xxx.lanl.gov/abs/physics/9907017
>
> > 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
>
> Bruce. How do you like your new test by Schutz?
I'm enjoying it. Especially the historical stuff that I wasn't familar
with. I'm pretty slow and I'm also trying to work on improving my math
abilities.
Bruce
Tom Roberts
> [someone] might conclude that GR predicts proper clock rates are
> coordinate-dependent.
But _all_ rates are coordinate dependent -- that's what we mean by
"rate". That is, the rate of K is dK/dt, and that only makes sense for a
time coordinate t. You can choose to measure this rate using a
collocated standard clock, in which case you are really using a time
coordinate determined by that clock. That is, one cannot differentiate
with respect to readings on a clock, only with respect to some
coordinate. But you can arrange for that coordinate to track the
readings of the clock.
I don't think we are in serious disagreement here, just terminology.
Tom Roberts tjro...@lucent.com
Tom Roberts
> As measured from a *single* coordinate system all clocks run at
> the same rate.
This is just plain not true[#]. For example, in the r>2M region of
Schwarzschild spacetime the rate of a standard clock (at rest) varies
depending on its radius. Here "rate" is defined relative to the usual
Schw. coordinates, so it is indeed "measured from a *single* coordinate
system".
[#] except for the very special case of coordinates in which
g_tt=1 everywhere.
At base, you are confusing yourself with unacknowledged puns on the word
"clock".
Say, rather, that all coordinate clocks tick at the same rate (for a
given coordinate system), but standard clocks tick at the rate of a
collocated coordinate clock only if the coordinates are appropriately
selected so that is so. Coordinate clocks can tick at any rate the mind
can imagine; standard clocks tick only at their standard _proper_ rate.
Note the coordinate clocks of Schw. coordinates are not
standard. They can be imagined thus: there is a standard clock
at spatial infinity that emits a light pulse at every tick;
each coordinate clock in the region r>2M ticks when those
light pulses reach it. There is no simple description for the
region r<2M (where the Schw. time coordinate is labeled "r").
Note that coordinate time is integrable over a region, while the proper
time of standard clocks is not, in general. This is why we need both,
and _must_ keep them separate. Precision in both thought and description
is absolutely essential, and your unacknowledged PUNs destroy that....
Tom Roberts tjro...@lucent.com
Sue...
Tom Roberts wrote:
> Pmb wrote:
> > As measured from a *single* coordinate system all clocks run at
> > the same rate.
>
> This is just plain not true[#]. For example, in the r>2M region of
> Schwarzschild spacetime the rate of a standard clock (at rest) varies
> depending on its radius. Here "rate" is defined relative to the usual
> Schw. coordinates, so it is indeed "measured from a *single* coordinate
> system".
>
> [#] except for the very special case of coordinates in which
> g_tt=1 everywhere.
>
> At base, you are confusing yourself with unacknowledged puns on the word
> "clock".
Try replacing the Schwarzschild *clock* with a vibrating reed
accelerometer. The 'Q' or stored energy leaks away when
the gravitational field of the oscillating mass becomes
non spherical due to the proximity of a large mass like a
planet.
Then you have something similar to the nuclear resonance or
hyperfine transititon of an atomic oscillator and similar behavior,
above and below, the surface of the earth.
http://www.research.ibm.com/grape/grape_ewald.htm
http://www.esa.int/SPECIALS/GSP/SEM0L6OVGJE_0.html
http://www.chem.purdue.edu/gchelp/liquids/inddip.html
Sue...
brian a m stuckless
> va...@cox.net wrote:
> > [someone] might conclude that GR predicts proper clock rates
> > are coordinate-dependent.
>
> But _all_ rates are coordinate dependent -- that's what we mean
> by "rate". That is, the rate of K is dK/dt, and that only makes
> sense for a time coordinate t. You can choose to measure this
> rate using a collocated standard clock, in which case you are
> really using a time coordinate determined by that clock.
>
> That is, one cannot differentiate
> with respect to readings on a clock,
$$ Tom you know GPS scientists use SEVERAL-clock's _average_ t.
$$ THERE would be NO NEED of SEVERAL-clocks, if you were RiGHT.
$$
$$ [The MORE atomic-TiCKERs ..the BETTER "clocking"-AVERAGE t].
$$ [MORE atomic-TiCs/STANDARD TiC; BETTER clocking _ACCURACY_].
$$ [Our STANDARD TiC-TOC TiME is STiLL Least-Squares-adjusted].
$$
$$ [TiME is the POiNT where the clock sits on it's WORLD-line].
$$ [TiMiNG is the *DiFFERENTiATiON* of DiSTiNGUiSHABLE POiNTs].
$$ [ -- A *MAGNiTUDE*-of-a-TiMiNG is WHAT a clock "READs" -- ].
$$
$$ COMPULSORY selection; ARBiTRARY choice.
$$ [Clock-READiNG: An UNadjusted MAGNiTUDE-of-TiMiNG counted ].
$$ [ANY "clock" is a *totally _ARBiTRARY_* human construction].
$$ [The clock-"READiNG" is ALSO a totally _ARBiTRARY _ choice].
$$
$$ [ i.e. NATURE does NOT mind if we have COHERENT STANDARDs ].
$$
$$ And the ARBiTRARiNESS is "of _CHOiCE_" not _SELECTiON_, duh.
$$ [ i.e. The CHOiCE is ARBiTRARY ..a SELECTiON is COMPULSORY].
$$
$$ You ARBiTRARiLY CHOOSE a COMPULSORY STANDARD ..then MEASURE.
> -=- only with respect to some coordinate.
> But you can arrange for that coordinate to
> track the readings of the clock.
>
> I don't think we are in serious disagreement here, just
> terminology.
$$ [ALPHABET is ARBiTRARY but we SELECT our CHOiCE, to WRiTE].
$$
$$ And "_all_ rates" areN'T just functions-of-TiMiNG ..dimwit.
> Tom [ He between his error-bars ] Roberts tjro...@lucent.com
Re: COMPULSORY selection; ARBiTRARY choice ..of the STANDARD set.
Re: NATURE doesN'T mind if we have COHERENT STANDARDs.
Re: GR frequency shift formula. ..End of POST.
brian a m stuckless
ululated:
> FuCkInG WhOrE DuMbBeLl tArT SlUtTy bItCh gAnG-RaPe bElLy sLaPpEr dUdEs
> $$ ToM YoU KnOw gPs fUcKiNg sOlIdEr sCiEnTiStS UsE SeVeRaL-ClOcK'S
> _aVeRaGe_ T. fUcKiNg bItChEn $$ ThErE WoUlD Be nO FuCkInG GuTtEr sLuT
> BiTcH NeEd oF SeVeRaL-ClOcKs, If yOu fUcKiNg bUm bRiStOlS WeRe rIgHt.
> $$ $$ [tHe fUcKiNg cOoN ArSe-fArMeR MoRe aToMiC-TiCkErS ..ThE BeTtEr
> "ClOcKiNg"-AvErAgE FuCkInG BuNg hOlE DrOp yOuR LoAd bItChErY T]. $$
> [MoRe aToMiC-TiCs/sTaNdArD TiC; fUcKiNg mOlL BuTtHeAd bEtTeR ClOcKiNg
> _AcCuRaCy_]. $$ [OuR FuCkInG ArSe-gUnNeR StAnDaRd tIc-tOc tImE Is
> sTiLl fUcKiNg bAlL BuStEr bUtTfAcE HaRlOt sHiTfIrE
> LeAsT-SqUaReS-AdJuStEd]. $$ $$ [TiMe iS FuCkInG CoCk tEaSeR FuCk
> hArLoT ThE PoInT WhErE ThE ClOcK FuCkInG AnAl-iNtRuDeR DuMb aSs sItS
> On iT'S WoRlD-LiNe]. $$ FuCkInG LeT ThE NiGgEr gIrL Do iT BlOwJoB ArSe
> pRiNcEsS ArSe-bEnDeR BuLlShIt aRtIsT [tImInG Is tHe *DiFfErEnTiAtIoN*
> oF FuCkInG FuCkHeAd fUcK Me dIsTiNgUiShAbLe pOiNtS]. $$ [ -- FuCkInG
> CoOn bRiStOlS A *mAgNiTuDe*-Of-a-tImInG Is wHaT A FuCkInG BeLlY WaRmEr
> cLoCk "ReAdS" -- ]. $$ FuCkInG WhAt tHe fUcK CoCkY NiGgEr $$
> FuCkInG BiTcH fUcKiNg pEcKeRhEaD CoMpUlSoRy sElEcTiOn; ArBiTrArY
> ChOiCe. $$ fUcKiNg pEnIs bReAtH [cLoCk-rEaDiNg: An uNaDjUsTeD
> MaGnItUdE-Of-tImInG CoUnTeD FuCkInG PuSsY LiPs bLaCk jOcK ]. $$ [aNy
> "ClOcK" iS FuCkInG BuTtHeAd bLoWjOb lItTlE SiStEr a *ToTaLlY
> _aRbItRaRy_* hUmAn cOnStRuCtIoN]. FuCkInG FuCk aRsEkIsSeR $$ [ThE
> ClOcK-"rEaDiNg" Is aLsO FuCkInG KrAuT A ToTaLlY _aRbItRaRy _ cHoIcE].
> FuCkInG BiTcH AsS $$ $$ [ I.E. nAtUrE FuCkInG SlUt dOeS NoT MiNd iF We
> fUcKiNg bItCh's bAsTaRd sHiThEaD HaVe cOhErEnT StAnDaRdS ]. $$ FuCkInG
> BrEaStIcLeS $$ aNd tHe aRbItRaRiNeSs iS FuCkInG BiTcH CoCkSuCkErS "oF
> _cHoIcE_" NoT _sElEcTiOn_, dUh. FuCkInG ShOoT YoUr wAd $$ [ I.E. tHe
> cHoIcE FuCkInG SmArTaLeCk nIgGeR MiCk nIgGeR LoVeR PrOsTiTuTe iS
> ArBiTrArY ..A SeLeCtIoN Is fUcKiNg jAcKoFf cOmPuLsOrY]. $$ $$ YoU
> ArBiTrArIlY FuCkInG BuTtFaCe cHoOsE A CoMpUlSoRy sTaNdArD ..ThEn
> fUcKiNg gOdDaMmIt cUm sTaIn mEaSuRe. BoNeR CuNt nIgGeR CuNt sCrUbBeR
> GaY DyKe wHoRe rApE I'D Do yOu cUnT StAb
> FuCkInG NiGgEr bAlL $$ [AlPhAbEt iS ArBiTrArY BuT FuCkInG BuTt aNaL
> JoUsTeR AnAl-bUcCaNeEr mR BaT ShIt wE SeLeCt oUr cHoIcE, tO FuCkInG
> BiSeXuAl bRoWn eYe wRiTe]. $$ $$ aNd "_aLl_ FuCkInG BoNe aChE ClIt
> cRoTcH BoYs sOn oF A WhOrE BaLlS DiCkSuCkEr nYmPhO ErEcTiOn
> aNaL-JaBbEr rAtEs" ArEn't jUsT FuNcTiOnS-Of-tImInG ..DiMwIt. FuCkInG
> SlUt cUnT-StIcKeR FuCk pOrK SwOrD HyMeN HaMmEr mUrDeR HoMo
> FuCkInG ShLoNg cHiNk wHoRe bUnG HoLe pUsSyLiPs bItCh rE: cOmPuLsOrY
> SeLeCtIoN; aRbItRaRy cHoIcE FuCkInG BaD ShIt fUcKuP ..Of tHe sTaNdArD
> SeT. rE: fUcKiNg cUnT BiTcH NaTuRe dOeSn't mInD If wE FuCkInG
> ShIt-fAcE HaVe cOhErEnT StAnDaRdS. rE: gR FuCkInG PiSsInG TiTtIeS BoNe
> aChE GoAt fUcKeR ArSe-fArMeR FrEqUeNcY ShIfT FoRmUlA. fUcKiNg
> aNaL-BoReR GoOk bItCh kIkE TwAt tArT ToE JaM CuNt sLaP SkIn
> FuCkInG ClIt FuCkInG AfRaId oF ThE NiGgErS CoCkSuCkEr
> FuCkInG CuM ..EnD Of pOsT. fUcKiNg pUsSyLiPs CuNt-sTiRrEr fLaPs
> cUnT-StAbBeR RiMjOb sTaRfIsH ShIt hEaD I'D Do yOu 69 DyKe lIkE It
> dIrTy
mlut...@wanadoo.fr
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.
brian a m stuckless
> >
> > So it doesn't matter what the definition of constant
> > velocity is, to you, right? You will define it and use
> > any way you want to, right? After all, it's a free
> > country and you can make up anything you want, right?
>
> There is only one definition dv/dt = 0
$$ Where, dv/t = x ..and, v/dt = y.
[ SPEED is ANY PATH/duration ; VELOCiTY the VECTOR/duration. ]
$$ You need BOTH velocity vector END-POiNT positions at ONCE.
$$ The VELOCiTY is ALWAYs known _ONLY_ in retrospect, so 'dv'
$$ caN'T EVER be known, at ONCE (i.e. ..at ANY *SiNGLE* 't').
$$ Atomic giga-frequency counters of REAL time.
The BEST atomic clock is SiMPLY a GiGA-frequency-COUNTER.!!
And, even the VERY best ATOMiC clocks CANNOT be SYNCRONiZED
TOGETHER, at once, on any GR or any OTHER sort of Ph.Tivity
TiME-line (i.e. WORLD-line), in ANY SPACE-time-continuum.!!
$$ All "clocks" are simply various FREQUENCY-COUNTERs used
even since very early on in PRE-history (to wit, period
tally-marks ..on excavated sticks & bones), to "TRY-to"
*CLOCK* ..the "REAL TiME".!! NOT to mention SUN & MOON.
However, a sort of TiME-AVERAGED simultaneity is achievable
..all having to compensate (REset) for design mysteries.!!
$$ DELTA v, (dv), caN'T *HAPPEN* ..all at ONCE.
$$ Simultaneity happens AT ONCE; EveryTHiNG else needs TiMiNG.
$$ And, *REAL* TiME is why *SHOP (LAB)* TiME has *ERROR* bars.
$$ ANY clock MOViNG ..or MOVED any LOWER ..will run FASTER.!!
$$ ANY clock MOViNG ..or MOVED any HiGHER, will run SLOWER.!!
$$ ANY clock's HORiZONTAL "SLOW transport", is UP transport.
$$ ANY clock moving OUT [ HORZ ] on a turn table", moves UP.
$$ ANY clock VARiES in ALTiTUDE for ANY horizontal TRANsPORT.
$$ Three VELOCiTY vector END-POiNT positions REQUiRED, for dv.
$$ Three VELOCiTY vector END-POiNTs (3 positions) define a dv.
$$ [GR Tivity CRACKED-pots: NOTE 'dt' caN'T HAPPEN ..at ONCE].
$$ Sam Wormly ..delta t (dt), PERiOD.
"It takes an infinite amount of time ..at which point,..".
-- Sam [ -worms ] Wormley.
>><> >><> >><> >><> >><>
$$ [2] Delta t (dt), PERiOD.
$$
$$ Tom [ He between his error-bars ] Roberts [@GR.BUFFY.com] wrote:
> va...@cox.net wrote:
> > [someone] might conclude that GR predicts proper clock rates
> > are coordinate-dependent.
>
Re: Delta t (dt), PERiOD. ..End of 2 rePOSTs.
va...@cox.net
mlut...@wanadoo.fr wrote:
> mlut...@wanadoo.fr wrote:
> > Koobee Wublee wrote:
> > > <mlut...@wanadoo.fr> wrote in message
> > > news:1143818636.6...@t31g2000cwb.googlegroups.com...
> > > >
> > > > What is the GR formula giving the frequency shift for the following
> > > > case:
> > > >
> > > > Two non-rotating spherical celestial objects 1 and 2 are orbiting along
> > > > their common center of gravity.
> > >
> > > which can predict all the following.
> > >
> > > ** Gravitational red shift through time dilation, g_00
> > > ** Gravitational blue shift through radial wavelength, 1 / g_00
> > > ** Gravitational no shift through tangential wavelength, 1
> > >
> > > Where
> > >
> > > ** ds^2 = c^2 g_00 dt^2 - dr^2 / g_00 - r^2 dH^2
> > >
> > > So, it depends on the experimental data. Whatever it comes out with,
> > > GR can predict it. On a historical note, gravitational red shift was
> > > predicted by Einstein after he declared the Equivalence Principle. It
> > > is nothing more than an event when Einstein finally understood gravity
> > > = acceleration and acceleration = gravity just as Newton had said many
> > > centuries ago. From his derivation of Doppler shift after peeking at
> > > Voigt's paper, Einstein somehow decided v^2 is sort of the same as
> > > acceleration. He thus decided gravitational acceleration as giving the
> > > same effect as if the source of gravitation (the sun, the planet, etc.)
> > > is moving away. Galilean mechanics would tell you that velocity is not
> > > the same as acceleration. This is as basic in the study of physics as
> > > it can get, and yet Einstein screwed it up.
> > >
> > > While the faithful believers of GR are patting on each other's back for
> > > the general belief, let's be a little more scientific about addressing
> > > this issue. As you know, energy must be conserved. As Planck has
> > > pointed out which Einstein re-iterated,
> > >
> > > E = h f
> > >
> > > A photon must retain its energy. Yet, Einstein's wrong analysis
> > > actually gave the correct result from various modern experiments.
> > > While the faithful would argue for Einstein's gifted intuition in this
> > > case, something else must be at work here. The true cause of
> > > gravitational red shift is still not yet discovered.
> > >
> > > Also, according to Lorentz Transform and thus Special Relavity which is
> > > just an interpretation to the mathematics of Lorentz Transform itself,
> > > your scenario with identical binary stars would certainly indicate blue
> > > shift. Both stars (their surface) have the same gravitational
> > > potential. Thus, gravity should play no role in this. However,
> > > claiming a blue shift would certainly not be in agreement with actual
> > > experimentation done if possibe.
> >
> > Thank you!
> >
> > 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 ?
> > Notice that the stars have different masses in my scenario.
> >
> > Marcel Luttgens
>
> Here is my approximate solution of the following problem.
> Do GR experts agree?
>
> 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?
>
> R1 = 5 Km,
> M2 = 5 M*,
> R2 = 20 km,
> d = 50 km.
>
> M1 = 1.989E+33 'g
> M2 = 1.989E+33 * 5 'g
> R1 = 500000! 'cm
> R2 = 2000000! 'cm
> d = 5000000! 'cm
>
> G = 6.673E-08 'Universal gravitational constant (CGS)
> c = 2.998E+10 'Speed of light in cm/s
>
> M1 = G * M1 / c ^ 2
> M2 = G * M2 / c ^ 2
>
> Nu2/Nu1 = 1 - M1*(1/R-1/(d-R2)-1/(2 * d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
> = 0.9
>
> In this case, one has a redshift!
As Tom Roberts pointed out 'the Marcel formula' is completely useless
in the strong field. To start with object 2 wouldn't be found in this
universe and object 1 most likely wouldn't exist either. Still you
could speculate 'what the answer would be' if they did exist according
to the parameters you set while disregarding system [ - dE/dt ] and
[ - dr/dt ] due to gravitational radiation. Something like this:
In geometric units
setting
1 = *, 2 = **
dt_shell** / dt_shell* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M*/r* )^1/2
= .511370707 / .337342555 = 1.515879625
For your case it's a blueshift.
Bruce
>
> Marcel Luttgens
va...@cox.net
Thanks for the clarification Tom. I'll work at the terminology.
Bruce
>
>
> Tom Roberts tjro...@lucent.com
va...@cox.net
Correction: For your case it's redshifted.
> Bruce
>
> >
> > Marcel Luttgens
mlut...@wanadoo.fr
Summary:
The GR formula giving the shift of a light signal of
frequency Nu0 emitted from the center of an object of mass M and
radius R, wrt the frequency Nu1 observed at the surface, is
Nu1/Nu0 = (3 * X - 1) / (2 * X), where
X = SQR(1 - 2 * M / R) and M = G * M / c ^ 2
Shift = Nu1/Nu0 - 1
The approximate GR formula is
Shift = 1 - GM/2R
Even for an object having the Sun mass and the Earth radius,
the approximate GR formula and the exact GR formula
give close results:
Shift (approx) = -1.1576E-4 and shift (exact) = -1.1580E-4,
meaning that generally, one can safely use approximate
formulae. Let's notice that such formulae can straightforwardly
by derived from the gain or loss of potential energy of photons
of pseudo-mass hNu/c^2.
Here is another problem for GR experts:
When a light of frequency N0 is emitted at a distance Rearth/2 from the
Earth center, what is the shift observed at the Earth surface?
According to a non-GR approximate formula
shift = - (G * Mearth / (Rearth ^ 2 * c ^ 2)) * 3 * Rearth / 8
= -2.61E-10
Imo, it would be very difficult, if not impossible, to derive
such formula from GR.
But perhaps the GR experts are more optimistic?
Anyhow, I don't expect an answer from the so-called GR expert who could
abuse, but not even find the GR formula Nu1/Nu0 = (3 * X - 1) / (2 *
X).
The other experts will probably avoid recognizing the inadequacies of
GR.
va...@cox.net
exhibited by cranks such as yourself.
Marcel Luttgen said:
"Here is my approximate solution of the following problem.
Do GR experts agree?
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?
Example: M1 = M* ( M* = 1 solar mass = 1.989E+33 g),
R1 = 5 Km,
M2 = 5 M*,
R2 = 20 km,
d = 50 km.
M1 = 1.989E+33 'g
M2 = 1.989E+33 * 5 'g
R1 = 500000! 'cm
R2 = 2000000! 'cm
d = 5000000! 'cm
G = 6.673E-08 'Universal gravitational constant (CGS)
c = 2.998E+10 'Speed of light in cm/s
M1 = G * M1 / c ^ 2
M2 = G * M2 / c ^ 2
Nu2/Nu1 = 1 - M1*(1/R-1/(d-R2)-1/(2 * d)) + M2*(1/R2-1/(d-R1)-1/(2*d))
= 0.9
In this case, one has a red shift!"
End of Marcel said.
It's been pointed out to Marcel that his formula will not work for
strong gravitational fields. The reason being his formula doesn't
account for relativistic effects. It's interesting to note the
relationship between Marcel's M2 and M1 are similar to the relationship
between the Earth and Moon. ie the center of system mass [barycenter]
is inside the radius of M2 just as the barycenter of the Earth-Moon
system is inside the radius of the Earth. Also note that M2 and M1 are
fictitious objects with exceedingly small probability of existing in
this universe. But if they could....
In geometric units:
r_barycenter = r_total [ M1/(M2 + M1) ], r_total = d.
= 50,000m [ 1477m / (7385m + 1477m) ] = 8333.333m from the center of
M2.
As M1 orbits the barycenter we can favorably compare the orbit of M2,
around barycenter, to the orbit of Earth around the Earth-Moon
barycenter. To find a reasonable approximation for predicting delta
period wrt a pulse of light emitted on the surface of M1 and received
on the surface of M2 the key parameters become the positions in the
gravitational field where the pulse of light is emitted and received.
Since both M1 and M2 are spherically symmetric and non-rotating we can
make the approximation using the Schwarzschild metric.
M2 velocity as it orbits the barycenter is exceedingly small compared
to the velocity of M1 as it orbits the barycenter.
Setting M2 = M**, M1 = M*, r2 = r**, r1 = r*, distance center of M1 to
barycenter = r***
v shell_M* = [ M** / ( r*** -2M**) ] ^1/2 = [ 7385m / (41667m -
14770m) ] ^1/2 = .524c
For M** we can use the first component of the Schwarzschild metric:
dtshell_M** = ( 1 - 2M**/r** ) ^1/2 dt, where dt = the GR Schwarzschild
bookkeeper coordinate time measured in flat spacetime at infinity
[boundary condition].
For M* we can use a formula derived from the Schwarzschild metric and
the effective potential of GR's equation of motion:
dtshell_M* = ( 1 - 3M**/r*** ) ^1/2 dt
____________________________________________________
Derivation:
Starting with the 2 dimensional form of the Schwarzschild metric and
setting dr = 0:
dTau^2 = (1 - 2M/r)dt^2 - 0 - r^2 dphi^2
Determining a useful expression for dphi:
L/m = r^2 (dphi/dTau)
dphi = [(L/m)/r^2]dTau
Substituting dphi^2 the metric becomes:
dTau^2 = ( 1 - 2M/r)dt^2 - [(L/m)^2/r^2]dTau^2
Dividing through by dt^2 will give the ratio (dTau/dt)^2:
(dTau/dt)^2 = (1 - 2M/r) - [(L/m)/r]^2(dTau/dt)^2
Simplified:
(dTau/dt)^2 = (1 - 2M/r) / [1 + (L/m)^2/r^2]
Using the equation of motion:
(dr/dTau)^2 = (E/m)^2 - (1 - 2M/r)[1 + (L/m)^2/r^2]
and taking the derivative of the effective potential wrt r, dividing
through by r^4 to express dV/dr as a quadratic, solving for 0, then r^2
can be expressed:
r^2 = [(L/m)^2]r - 3(L/m)^2
Substituting [(L/m)^2]r - 3(L/m)^2 for r^2 into (dTau/dt)^2 = [..]
results in:
(dTau/dt)^2=(1 - 2M/r)/[1 + (L/m)^2)/[(L/m)^2)r - 3(L/m)^2]
Simplified:
dTau/dt = (1 - 3M/r)^1/2
_______________________________________________________________
Bact to Marcel's folly:
To find the ratio [ dtshell_M** / dtshell_M* ]
dtshell_M* / ( 1 - 3M**/r*** ) ^1/2 = dt = dtshell_M** / ( 1 -
2M**/r** )^1/2
dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2
= .511370707 / .639687422 = .74727606
Throughout this thread Marcel has claimed GR isn't a useful theory for
doing relativistic physics. Since he knows very little about GR it's
difficult to understand how he came to such a conclusion. It is well
understood that weak field approximations are useless for describing
strong field physics. Apparently well understood by everybody but
Marcel Luttgen.
va...@cox.net
Correction: ".........., multiply through by r^4 to express dV/dr as a
quadratic, ........:
brian a m stuckless
>><> >><> >><> >><> >><>
> > [someone] might conclude that GR predicts proper clock rates
> > are coordinate-dependent.
>
> But _all_ rates are coordinate dependent -- that's what we mean
> by "rate". That is, the rate of K is dK/dt, and that only makes
> sense for a time coordinate t. You can choose to measure this
> rate using a collocated standard clock, in which case you are
> really using a time coordinate determined by that clock.
>
> That is, one cannot differentiate
> with respect to readings on a clock,
$$ Tom you know GPS scientists use SEVERAL-clock's _average_ t.
$$ THERE would be NO NEED of SEVERAL-clocks, if you were RiGHT.
$$
$$ [The MORE atomic-TiCKERs ..the BETTER "clocking"-AVERAGE t].
$$ [MORE atomic-TiCs/STANDARD TiC; BETTER clocking _ACCURACY_].
$$ [Our STANDARD TiC-TOC TiME is STiLL Least-Squares-adjusted].
$$
$$ [TiME is the POiNT where the clock sits on it's WORLD-line].
$$ [TiMiNG is the *DiFFERENTiATiON* of DiSTiNGUiSHABLE POiNTs].
$$ [ -- A *MAGNiTUDE*-of-a-TiMiNG is WHAT a clock "READs" -- ].
$$
$$ COMPULSORY selection; ARBiTRARY choice.
$$ [Clock-READiNG: An UNadjusted MAGNiTUDE-of-TiMiNG counted ].
$$ [ANY "clock" is a *totally _ARBiTRARY_* human construction].
$$ [The clock-"READiNG" is ALSO a totally _ARBiTRARY _ choice].
$$
$$ [ i.e. NATURE does NOT mind if we have COHERENT STANDARDs ].
$$
$$ And the ARBiTRARiNESS is "of _CHOiCE_" not _SELECTiON_, duh.
$$ [ i.e. The CHOiCE is ARBiTRARY ..a SELECTiON is COMPULSORY].
$$
$$ You ARBiTRARiLY CHOOSE a COMPULSORY STANDARD ..then MEASURE.
> -=- only with respect to some coordinate.
> But you can arrange for that coordinate to
> track the readings of the clock.
>
> I don't think we are in serious disagreement here,
> just terminology.
$$ [ALPHABET is ARBiTRARY but we SELECT our CHOiCE, to WRiTE].
mlut...@wanadoo.fr
Thank you! I am having a closer look at your formula.
Marcel Luttgens
Hexenmeister
<mlut...@wanadoo.fr> wrote in message news:1145784786....@u72g2000cwu.googlegroups.com...
| > > Throughout this thread Marcel has claimed GR isn't a useful theory for
| > > doing relativistic physics.
I don't see that. I'm sure GR is very useful for relativistic physics.
Pity relativistic physics has nothing to do with real physics, since
relativistic physics is only screwed up mathematics.
"Real" has nothing to do with it. - Tom Roberts
Androcles.
mlut...@wanadoo.fr
It accounts for relativistic effects, but part of them are ignored in
my
approximate formula.
> It's interesting to note the
> > relationship between Marcel's M2 and M1 are similar to the relationship
> > between the Earth and Moon. ie the center of system mass [barycenter]
> > is inside the radius of M2 just as the barycenter of the Earth-Moon
> > system is inside the radius of the Earth.
Yes, but I purposely chose the short distance of 50000 km to get clear
relativistic effects.
> Also note that M2 and M1 are
> > fictitious objects with exceedingly small probability of existing in
> > this universe. But if they could....
Sure, but this is the fun of thought experiments.
> >
> > In geometric units:
> >
> > r_barycenter = r_total [ M1/(M2 + M1) ], r_total = d.
> >
> > = 50,000m [ 1477m / (7385m + 1477m) ] = 8333.333m from the center of
> > M2.
Yes.
> >
> > As M1 orbits the barycenter we can favorably compare the orbit of M2,
> > around barycenter, to the orbit of Earth around the Earth-Moon
> > barycenter. To find a reasonable approximation for predicting delta
> > period wrt a pulse of light emitted on the surface of M1 and received
> > on the surface of M2 the key parameters become the positions in the
> > gravitational field where the pulse of light is emitted and received.
> >
> > Since both M1 and M2 are spherically symmetric and non-rotating we can
> > make the approximation using the Schwarzschild metric.
Otherwise, the GR solution would have been much more difficult, if not
impossible, to derive.
> >
> > M2 velocity as it orbits the barycenter is exceedingly small compared
> > to the velocity of M1 as it orbits the barycenter.
It is not "extremely" small. You should take it into account.
> >
> > Setting M2 = M**, M1 = M*, r2 = r**, r1 = r*, distance center of M1 to
> > barycenter = r***
> >
> > v shell_M* = [ M** / ( r*** -2M**) ] ^1/2 = [ 7385m / (41667m -
> > 14770m) ] ^1/2 = .524c
Are you sure of your formula? I found .35 c.
Would you use the same formula to get the orbital velocity of the Earth
around
the Sun?
Here you are a little nasty! But thank you nevertheless.
> >
> > To find the ratio [ dtshell_M** / dtshell_M* ]
> >
> > dtshell_M* / ( 1 - 3M**/r*** ) ^1/2 = dt = dtshell_M** / ( 1 -
> > 2M**/r** )^1/2
> >
> >
> > dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2
> >
Could you give your solution in terms of M* and M** ?
Don't forget that the pulse is emitted from the surface of M1. Where is
R1 (your r*)
in your formula?
> >
> >
> > = .511370707 / .639687422 = .74727606
> >
> > Throughout this thread Marcel has claimed GR isn't a useful theory for
> > doing relativistic physics. Since he knows very little about GR it's
> > difficult to understand how he came to such a conclusion. It is well
> > understood that weak field approximations are useless for describing
> > strong field physics. Apparently well understood by everybody but
> > Marcel Luttgen.
I am skeptical about your elaborations.
Marcel Luttgens
mlut...@wanadoo.fr
A simple way to prove that your formula is false:
Let M* = M** = 7383.5 m
r* = r** = 20000 m
d = 200 000 000 m (thus r*** = 100 000 000 m)
As d is so big, your formula reduces to
dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 = .511,
instead of a value very close to 1.
(the relativistic effects are negligible).
Marcel Luttgens
va...@cox.net
You set d = 50km = 50,000m.
The formula you just wrote doesn't make any sense. The reason is you
don't understand what dtshell represents.
dtshell = dTau
dt = Schwarzschild bookkeeper coordinate time measured in flat
spacetime far away [ie at infinity]. The boundary condition. The
formula is the first component of the metric.
dtshell = ( 1 - 2M/r )^1/2 dt
The ratio would be:
dtshell/dt = ( 1 - 2M/r )^1/2
For the ratio [dtshell_M** / dtshell_M*]:
dtshell_M**/( 1 - 2M**/r** )^1/2 = dt = dtshell_M*/( 1 - 2M*/r* )^1/2
So:
[dtshell_M** / dtshell_M*] = ( 1 - 2M**/r** )^1/2 / ( 1 - 2M*/r* )^1/2
= .511371 / .6396874
= .799408
My approximation acknowledged that M* is essentially a satellite of
M**. The satellite shell velocity [measured with a clock on the shell]
is .524c. The derivation resulting in dTau/dt = ( 1 - 3M/r )^1/2
accounts for that. So the final ratio becomes
dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2
= .511370707 / .639687422 = .74727606
Even so; I agree with Tom that doing this type of approximation, for
strong field physics, isn't the best way to go. But at the least it
accounts for major strong field relativistic effects.
Essentially this is what you did:
dtshell/dt = ( 1 - 2M/r )^1/2
You approximated:
dtshell/dt = [1 - 2M/2r ] = 1 - M/r
dtshell_M** / dtshell_M* = [ 1 - M**/r** ] / [ 1 - M*/r* ] = .63075 /
.7046 = .8952
So why does this weak field approximation work so well for our solar
system [all weak gravitational fields].
For the Earth the ratio dtshell/dt becomes:
dtshell/dt = [1 - 2(.00444m)/6.371E6 ]^1/2 = .999999999 / 1 =
.999999999
Which is exceptionally close to the Newton prediction of 1.
Aside: The velocity you calculated appears to have been calculated
using the boundary clock with r = 50,000m [even then I get .3843c]. For
the Schwrazschild geometry that is referred to as the bookkeeper
velocity.
The thing I find disturbing, about this conversation, is your
willingness to discount GR as a viable tool for doing physics. It
hasn't failed an empirical test to date and I don't think you've
actually studied the theory. Hopefully this analysis will clarify why
the weak field approximation, and Newtonian gravitational physics,
doesn't work in the strong field.
Best wishes
Bruce
mlut...@wanadoo.fr
Remember:
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?
Your formula should apply to all cases, for instance
M1 = M* ( M* = 1 solar mass = 1.989E+33 g),
R1 = 5 Km,
M2 = M*,
R2 = 5 km,
d = 2 000 000 000 km.
The answer is obviously Nu2/Nu1 ~ 1.
I am claiming that, as your formula gives a value that is very far from
1,
it is false.
Marcel Luttgens
va...@cox.net
In the first scenerio M1 can be treated as a satellite of M2. The new
case neither M1 or M2 is a satellite of the other. Why should the
formula I used for the first case apply to all cases? The geometry of
the systems can be different.Your approximation doesn't apply to all
cases regardless what you think. For the new case I would use:
[dtshell_M** / dtshell_M*] = ( 1 - 2M**/r** )^1/2 / ( 1 - 2M*/r* )^1/2
= .6396874 / .6396874 = 1
Look at what I wrote earlier. Using the above formula the answer to
your first case was
.799408. I treated M1 as a satellite of M2 [because the geometry
proclaims it is a satellite] to include M1 shell velocity because it
definitely makes a difference [ie you can't ignore it]. Then the answer
became .74727606. Ignoring relativistic effects your answer was .9. I
showed you why there is such a difference and why weak field
approximations and Newtonian gravitational physics don't work for
strong field gravitational physics. Your claim is based in ignorance.
There really isn't any reason to go further. It was fun because I enjoy
this stuff.
Later
Bruce
mlut...@wanadoo.fr
The problem was finding a *general* GR solution for the following
scenario:
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?
What is your general formula?
Obviously not the one you presented, as it gives a wrong result
when M1=M2, R1=R2 and d big.
Recognize that you can't derive such general GR formula.
But if you can, there really is a good reason to go further.
Marcel Luttgens
mlut...@wanadoo.fr
reproduce the whole discussion:
The problem was finding a *general* GR solution for the following
scenario:
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?
What is your general formula?
Hexenmeister
<mlut...@wanadoo.fr> wrote in message news:1145885127.3...@y43g2000cwc.googlegroups.com...
| Here is again my last response, because it is not necessary to
| reproduce the whole discussion:
|
| The problem was finding a *general* GR solution for the following
| scenario:
|
| Two non-rotating spherical celestial objects 1 and 2 are orbiting along
|
| their common center of gravity.
"Barycentre" in English (or if American, "barycenter", it means the same).
http://en.wikipedia.org/wiki/Barycentre
| 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?
Actually an intriguing question since it isn't (in general) Nu1 by
Celestial Mechanics.
Unfortunately you haven't supplied sufficient information.
The problem is d, which is not constant for an elliptical orbit
and in the general case the orbit is elliptical. Thus there is a velocity
component along the line of separation and consequently
a frequency shift, sometimes red and sometimes blue.
Nu2 = Nu1 (c+v)/c
Androcles.
mlut...@wanadoo.fr
Moreover, from the data
d = 50000 m
M* = 1477 m
M** = 7385 m
r** = 20000 m
r*** = 41667 m
you got a frequency ratio of .74727606, with the help of your formula
dtshell_M** / dtshell_M*
= ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/r*** )^1/2
= .511370707 / .639687422 = .74727606
You claimed:
"I treated M1 as a satellite of M2 [because the geometry
proclaims it is a satellite] to include M1 shell velocity because it
definitely makes a difference [ie you can't ignore it]. Then the answer
became .74727606."
The fact that M1 is a satellite of M2 is independant from d.
As your r*** = d(1-M*/(M* + M**)) = .8333 d
your formula can be written
dtshell_M** / dtshell_M*
= ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/.8333 d )^1/2
So, if d is big, for instance 200 000 000 m, the denominator
( 1 - 3M**/.8333 d )^1/2 becomes ~1, and your formula reduces
to dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 =~ .511
This makes of course no sense.
You should have gotten 1.074 instead!
Your claim that your formula is valid when M1 can be treated
as a satellite is based on ignorance. :-)
Marcel Luttgens
va...@cox.net
I've shown you a formula for both scenerios. Each formula fits the
geometry of the case it's applied to. I've also shown you why your
formula doesn't work for the strong field. There is no reason to go
further.
Bruce
>
> Marcel Luttgens
mlut...@wanadoo.fr
va...@cox.net wrote:
> mlut...@wanadoo.fr wrote:
> > Here is again my last response, because it is not necessary to
> > reproduce the whole discussion:
> >
> > The problem was finding a *general* GR solution for the following
> > scenario:
> >
> > 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?
> >
> > What is your general formula?
I retain
- that you couldn't find a general GR solution for the above scenario.
- that your solution in which you treated M1 as a satellite of M2 leads
to wrong results, meaning that it is false.
Indeed, the fact that M1 is a satellite of M2 is independant from d,
the
distance between the centers of M and M.
As your r*** = d(1-M*/(M* + M**)) = .8333 d, your formula can be
written
dtshell_M** / dtshell_M*
= ( 1 - 2M**/r** )^1/2 / ( 1 - 3M**/.8333 d )^1/2
So, if d is big, for instance 200 000 000 m, the denominator
( 1 - 3M**/.8333 d )^1/2 becomes ~1, and your formula reduces
to dtshell_M** / dtshell_M* = ( 1 - 2M**/r** )^1/2 =~ .511
This makes of course no sense.
You should have gotten 1.074 instead, a blue shift, not an enormous
redshit.
-that the derivation of your second solution has not be presented
Yes, you have shown a formula for both scenarios, but the first one is
definitvely false, and the second one seems ad hoc.
There are very good reasons to go further, but by somebody who has a
very good
knowledge of GR theory, and also a good practice of the mathematical
verification of the formulae he found.
Marcel Luttgens