| Topic: |
Science > Physics |
| User: |
"George Hammond" |
| Date: |
14 Sep 2005 10:10:19 AM |
| Object: |
Re: Newton's from Einstein? |
On 14 Sep 2005 02:01:11 -0700, "Eric Gisse" <jowr.pi@gmail.com> wrote:
George Hammond wrote:
[snip]
George, are you mixing it up so we don't always expect you to be
talking about crazy *****?
[Hammond]
You're the one who is mixed up Wombat, not me.
===restore oringinal post=========
George Hammond" <nowhere1@nospam.org> wrote in message
news:6qsci15dnc10km75t8o6okt5n1hjpj4v5f@4ax.com...
On Tue, 13 Sep 2005 04:26:53 GMT, David Cross <spamdenied@nospam.net>
wrote:
Hi,
I'm sitting in on a General Relativity course and so we're going through
all the sodding tensor stuff which I haven't much knowledge of, but
am attempting to wrap my head around. :)
Anyway, I was curious. Just for kicks, how does one derive the
familiar F= Gm1 m2 / r^2 equation from the Einstein tensor form?
---
David Cross
[Hammond]
The "inverse square law" refers to the force due to a spherical
mass body.
The case of a static spherical mass body can be solved exactly
in General Relativity and is known as the "Schwarzchild
Solution".
From the Schwarzchild solution for the metric, one can
easily compute the Christoffel Symbols.
The Christoffel Symbols appear in the "geodesic equation",
which is the equation of motion in General Relativity.
For a static spherical mass the geodetic equation reduces
to:
d^2 r/dt^2 = - M/r^2 (1 - 2M/r)
This is the EXACT General Relativity solution for pure radial
force. For ordinary masses (less than black holes) 1 >> 2M/r
therefore the solution is "approximately" given by:
F = ma = d^2 r / dt^2 = -M/r^2 which is Newton's familiar "inverse
square force law"
George is of course correct. But it can also be found in any book on
GR where they determine the constants k and c in the EFE's
kG + c = T by taking the non relativistic limit and getting Newton's
equation. Have a look at
http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html
under gravitation and from equation 4.4.
Thanks
Bill
[Hammond]
Sure... the lengthy "linearized analysis" is presented in
every textbook.... but it isn't a "simple answeer to a simple
question"........ namely: How does one derive Newton's
inverse square law from Einstein's equations.... which is what
the guy asked.... and is what every beginning student wants to
know...since it is obviously the "first question" anyone would
ask.
Therefore, I have presented the "simple answer" to the
"simple question" above. And frankly, I've NEVER even seen it
in a textbook!
My simple derivation, valid for the static case (which is what
everyone is interested in), shows clearly the difference
(the correction term) to Newton's inverse square law that comes
from General Relativity. In fact it explains why they actually
proposed at one time to alter Newton's inverse square law
in the "sixteenth decimal place" to account for the advance of
the perihelion of Mercury before GR was discovered.
========================================
SCIENTIFIC PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
=======================================
Join COSA church (Church of the Scientific Advent)
Send a blank email to
and your email address will be added to the
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===========================
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|
|
| User: "John C. Polasek" |
|
| Title: Re: Newton's from Einstein? |
17 Sep 2005 10:53:51 AM |
|
|
On Wed, 14 Sep 2005 15:10:19 GMT, George Hammond <nowhere1@nospam.org>
wrote:
On 14 Sep 2005 02:01:11 -0700, "Eric Gisse" <jowr.pi@gmail.com> wrote:
George Hammond wrote:
[snip]
George, are you mixing it up so we don't always expect you to be
talking about crazy *****?
[Hammond]
You're the one who is mixed up Wombat, not me.
===restore oringinal post=========
George Hammond" <nowhere1@nospam.org> wrote in message
news:6qsci15dnc10km75t8o6okt5n1hjpj4v5f@4ax.com...
On Tue, 13 Sep 2005 04:26:53 GMT, David Cross <spamdenied@nospam.net>
wrote:
Hi,
I'm sitting in on a General Relativity course and so we're going through
all the sodding tensor stuff which I haven't much knowledge of, but
am attempting to wrap my head around. :)
Anyway, I was curious. Just for kicks, how does one derive the
familiar F= Gm1 m2 / r^2 equation from the Einstein tensor form?
---
David Cross
[Hammond]
The "inverse square law" refers to the force due to a spherical
mass body.
The case of a static spherical mass body can be solved exactly
in General Relativity and is known as the "Schwarzchild
Solution".
From the Schwarzchild solution for the metric, one can
easily compute the Christoffel Symbols.
The Christoffel Symbols appear in the "geodesic equation",
which is the equation of motion in General Relativity.
For a static spherical mass the geodetic equation reduces
to:
d^2 r/dt^2 = - M/r^2 (1 - 2M/r)
This is the EXACT General Relativity solution for pure radial
force. For ordinary masses (less than black holes) 1 >> 2M/r
therefore the solution is "approximately" given by:
F = ma = d^2 r / dt^2 = -M/r^2 which is Newton's familiar "inverse
square force law"
George is of course correct.
Not so, see below for escape velocity calculation.JP
But it can also be found in any book on
GR where they determine the constants k and c in the EFE's
kG + c = T by taking the non relativistic limit and getting Newton's
equation. Have a look at
http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html
under gravitation and from equation 4.4.
Thanks
Bill
[Hammond]
Sure... the lengthy "linearized analysis" is presented in
every textbook.... but it isn't a "simple answeer to a simple
question"........ namely: How does one derive Newton's
inverse square law from Einstein's equations.... which is what
the guy asked.... and is what every beginning student wants to
know...since it is obviously the "first question" anyone would
ask.
Therefore, I have presented the "simple answer" to the
"simple question" above. And frankly, I've NEVER even seen it
in a textbook!
My simple derivation, valid for the static case (which is what
everyone is interested in), shows clearly the difference
(the correction term) to Newton's inverse square law that comes
from General Relativity. In fact it explains why they actually
proposed at one time to alter Newton's inverse square law
in the "sixteenth decimal place" to account for the advance of
the perihelion of Mercury before GR was discovered.
========================================
SCIENTIFIC PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
=======================================
Join COSA church (Church of the Scientific Advent)
Send a blank email to
and your email address will be added to the
COSA discussion list (free, no obligation)
===========================
and please ask your news service to add:
alt.sci.relativistic-proof-of-god.moderated
===========================
George, George, George, before we get all puffed up with simple
derivations and stuff (so far unpublished) I thought your result
looked wrong at first sight. (I lost sight of this message due to
some vicious pruning).
It is simple enough to test your expression by computing the escape
velocity where V^2/2 = MG/r
by taking the sqrt of the integral of your expression from infinity
down to Rearth
sqrt(Int( - M/r^2 (1 - 2M/r)))dr = 7903 m/s
Using Newton we get
sqrt(Int(MG/r^2)dr = 11,180 m/s
That is the universally acknowledged escape velocity, so unless I have
made an error this proves your GR expression for a law of gravity to
be incorrect.
John Polasek
http://www.dualspace.net
.
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Newton's from Einstein? |
17 Sep 2005 12:20:02 PM |
|
|
On Sat, 17 Sep 2005 15:53:51 GMT, John C. Polasek
<jpolasek@cfl.rr.com> wrote:
On Wed, 14 Sep 2005 15:10:19 GMT, George Hammond <nowhere1@nospam.org>
wrote:
On 14 Sep 2005 02:01:11 -0700, "Eric Gisse" <jowr.pi@gmail.com> wrote:
George Hammond wrote:
[snip]
George, are you mixing it up so we don't always expect you to be
talking about crazy *****?
[Hammond]
You're the one who is mixed up Wombat, not me.
===restore oringinal post=========
George Hammond" <nowhere1@nospam.org> wrote in message
news:6qsci15dnc10km75t8o6okt5n1hjpj4v5f@4ax.com...
On Tue, 13 Sep 2005 04:26:53 GMT, David Cross <spamdenied@nospam.net>
wrote:
Hi,
I'm sitting in on a General Relativity course and so we're going through
all the sodding tensor stuff which I haven't much knowledge of, but
am attempting to wrap my head around. :)
Anyway, I was curious. Just for kicks, how does one derive the
familiar F= Gm1 m2 / r^2 equation from the Einstein tensor form?
---
David Cross
[Hammond]
The "inverse square law" refers to the force due to a spherical
mass body.
The case of a static spherical mass body can be solved exactly
in General Relativity and is known as the "Schwarzchild
Solution".
From the Schwarzchild solution for the metric, one can
easily compute the Christoffel Symbols.
The Christoffel Symbols appear in the "geodesic equation",
which is the equation of motion in General Relativity.
For a static spherical mass the geodetic equation reduces
to:
d^2 r/dt^2 = - M/r^2 (1 - 2M/r)
This is the EXACT General Relativity solution for pure radial
force. For ordinary masses (less than black holes) 1 >> 2M/r
therefore the solution is "approximately" given by:
F = ma = d^2 r / dt^2 = -M/r^2 which is Newton's familiar "inverse
square force law"
George is of course correct.
Not so, see below for escape velocity calculation.JP
But it can also be found in any book on
GR where they determine the constants k and c in the EFE's
kG + c = T by taking the non relativistic limit and getting Newton's
equation. Have a look at
http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html
under gravitation and from equation 4.4.
Thanks
Bill
[Hammond]
Sure... the lengthy "linearized analysis" is presented in
every textbook.... but it isn't a "simple answeer to a simple
question"........ namely: How does one derive Newton's
inverse square law from Einstein's equations.... which is what
the guy asked.... and is what every beginning student wants to
know...since it is obviously the "first question" anyone would
ask.
Therefore, I have presented the "simple answer" to the
"simple question" above. And frankly, I've NEVER even seen it
in a textbook!
My simple derivation, valid for the static case (which is what
everyone is interested in), shows clearly the difference
(the correction term) to Newton's inverse square law that comes
from General Relativity. In fact it explains why they actually
proposed at one time to alter Newton's inverse square law
in the "sixteenth decimal place" to account for the advance of
the perihelion of Mercury before GR was discovered.
========================================
SCIENTIFIC PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
=======================================
Join COSA church (Church of the Scientific Advent)
Send a blank email to
and your email address will be added to the
COSA discussion list (free, no obligation)
===========================
and please ask your news service to add:
alt.sci.relativistic-proof-of-god.moderated
===========================
George, George, George, before we get all puffed up with simple
derivations and stuff (so far unpublished) I thought your result
looked wrong at first sight. (I lost sight of this message due to
some vicious pruning).
It is simple enough to test your expression by computing the escape
velocity where V^2/2 = MG/r
by taking the sqrt of the integral of your expression from infinity
down to Rearth
sqrt(Int( - M/r^2 (1 - 2M/r)))dr = 7903 m/s
Using Newton we get
sqrt(Int(MG/r^2)dr = 11,180 m/s
That is the universally acknowledged escape velocity, so unless I have
made an error this proves your GR expression for a law of gravity to
be incorrect.
John Polasek
http://www.dualspace.net
My apologies George, I forgot to divide V^2 by 2 and it's OK. But I
see a second term that comes to 0.208 m/s that is the redshift due to
earth's gravity so there is a slight effect at earth.
John Polasek
http://www.dualspace.net
.
|
|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Newton's from Einstein? |
14 Sep 2005 01:46:19 PM |
|
|
On Wed, 14 Sep 2005 15:10:19 GMT, George Hammond <nowhere1@nospam.org>
wrote:
On 14 Sep 2005 02:01:11 -0700, "Eric Gisse" <jowr.pi@gmail.com> wrote:
George Hammond wrote:
[snip]
George, are you mixing it up so we don't always expect you to be
talking about crazy *****?
[Hammond]
You're the one who is mixed up Wombat, not me.
===restore oringinal post=========
George Hammond" <nowhere1@nospam.org> wrote in message
news:6qsci15dnc10km75t8o6okt5n1hjpj4v5f@4ax.com...
On Tue, 13 Sep 2005 04:26:53 GMT, David Cross <spamdenied@nospam.net>
wrote:
Hi,
I'm sitting in on a General Relativity course and so we're going through
all the sodding tensor stuff which I haven't much knowledge of, but
am attempting to wrap my head around. :)
Anyway, I was curious. Just for kicks, how does one derive the
familiar F= Gm1 m2 / r^2 equation from the Einstein tensor form?
---
David Cross
[Hammond]
The "inverse square law" refers to the force due to a spherical
mass body.
The case of a static spherical mass body can be solved exactly
in General Relativity and is known as the "Schwarzchild
Solution".
From the Schwarzchild solution for the metric, one can
easily compute the Christoffel Symbols.
The Christoffel Symbols appear in the "geodesic equation",
which is the equation of motion in General Relativity.
For a static spherical mass the geodetic equation reduces
to:
d^2 r/dt^2 = - M/r^2 (1 - 2M/r)
This is the EXACT General Relativity solution for pure radial
force. For ordinary masses (less than black holes) 1 >> 2M/r
therefore the solution is "approximately" given by:
F = ma = d^2 r / dt^2 = -M/r^2 which is Newton's familiar "inverse
square force law"
George is of course correct. But it can also be found in any book on
GR where they determine the constants k and c in the EFE's
kG + c = T by taking the non relativistic limit and getting Newton's
equation. Have a look at
http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html
under gravitation and from equation 4.4.
Thanks
Bill
[Hammond]
Sure... the lengthy "linearized analysis" is presented in
every textbook.... but it isn't a "simple answeer to a simple
question"........ namely: How does one derive Newton's
inverse square law from Einstein's equations.... which is what
the guy asked.... and is what every beginning student wants to
know...since it is obviously the "first question" anyone would
ask.
Therefore, I have presented the "simple answer" to the
"simple question" above. And frankly, I've NEVER even seen it
in a textbook!
My simple derivation, valid for the static case (which is what
everyone is interested in), shows clearly the difference
(the correction term) to Newton's inverse square law that comes
from General Relativity. In fact it explains why they actually
proposed at one time to alter Newton's inverse square law
in the "sixteenth decimal place" to account for the advance of
the perihelion of Mercury before GR was discovered.
========================================
SCIENTIFIC PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
=======================================
Join COSA church (Church of the Scientific Advent)
Send a blank email to
and your email address will be added to the
COSA discussion list (free, no obligation)
===========================
and please ask your news service to add:
alt.sci.relativistic-proof-of-god.moderated
===========================
Besides not seeing how you get this from the Schwarzschild metric, you
have managed to use M twice as if they were the same, but couldn't
possibly be as you should write MG/r^2 and MG/c^2r instead of M/r^2
and 2M/r.
John Polasek
http://www.dualspace.net
.
|
|
|
| User: "George Hammond" |
|
| Title: Re: Newton's from Einstein? |
14 Sep 2005 08:20:07 PM |
|
|
On Wed, 14 Sep 2005 18:46:19 GMT, John C. Polasek
<jpolasek@cfl.rr.com> wrote:
On Wed, 14 Sep 2005 15:10:19 GMT, George Hammond <nowhere1@nospam.org>
wrote:
On 14 Sep 2005 02:01:11 -0700, "Eric Gisse" <jowr.pi@gmail.com> wrote:
George Hammond wrote:
[snip]
George, are you mixing it up so we don't always expect you to be
talking about crazy *****?
[Hammond]
You're the one who is mixed up Wombat, not me.
===restore oringinal post=========
George Hammond" <nowhere1@nospam.org> wrote in message
news:6qsci15dnc10km75t8o6okt5n1hjpj4v5f@4ax.com...
On Tue, 13 Sep 2005 04:26:53 GMT, David Cross <spamdenied@nospam.net>
wrote:
Hi,
I'm sitting in on a General Relativity course and so we're going through
all the sodding tensor stuff which I haven't much knowledge of, but
am attempting to wrap my head around. :)
Anyway, I was curious. Just for kicks, how does one derive the
familiar F= Gm1 m2 / r^2 equation from the Einstein tensor form?
---
David Cross
[Hammond]
The "inverse square law" refers to the force due to a spherical
mass body.
The case of a static spherical mass body can be solved exactly
in General Relativity and is known as the "Schwarzchild
Solution".
From the Schwarzchild solution for the metric, one can
easily compute the Christoffel Symbols.
The Christoffel Symbols appear in the "geodesic equation",
which is the equation of motion in General Relativity.
For a static spherical mass the geodetic equation reduces
to:
d^2 r/dt^2 = - M/r^2 (1 - 2M/r)
This is the EXACT General Relativity solution for pure radial
force. For ordinary masses (less than black holes) 1 >> 2M/r
therefore the solution is "approximately" given by:
F = ma = d^2 r / dt^2 = -M/r^2 which is Newton's familiar "inverse
square force law"
[Hobba]
George is of course correct. But it can also be found in any book on
GR where they determine the constants k and c in the EFE's
kG + c = T by taking the non relativistic limit and getting Newton's
equation. Have a look at
http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html
under gravitation and from equation 4.4.
Thanks
Bill
[Hammond]
Sure... the lengthy "linearized analysis" is presented in
every textbook.... but it isn't a "simple answeer to a simple
question"........ namely: How does one derive Newton's
inverse square law from Einstein's equations.... which is what
the guy asked.... and is what every beginning student wants to
know...since it is obviously the "first question" anyone would
ask.
Therefore, I have presented the "simple answer" to the
"simple question" above. And frankly, I've NEVER even seen it
in a textbook!
My simple derivation, valid for the static case (which is what
everyone is interested in), shows clearly the difference
(the correction term) to Newton's inverse square law that comes
from General Relativity. In fact it explains why they actually
proposed at one time to alter Newton's inverse square law
in the "sixteenth decimal place" to account for the advance of
the perihelion of Mercury before GR was discovered.
[Polasek]
Besides not seeing how you get this from the Schwarzschild metric,
[Hammond]
The geodesic equation for the Schwarzchild metric; the pure "static"
case (no movement of mass M or of test particle m) involves only
one Christiffel symbol, namely:
__r
| tt = (M/r^2)(1 - 2M/r)
The reason there is only one Christiffel symbol involved in the
equation of motion is because the term (dXa/ds)(dXb/ds) in the
geodesic equation is ZERO for all terms except dX4 = dt and then
dt/ds = 1 ... in the static case (NO movement).
This Christoffel symbol may be calculated directly from the metric
(see any textbook for the well known formula), or can be found listed
in the handy table given in Hartle, _Gravity_, Addison Wesley 2003
(p. 546), or any one of a number of texts.
you
have managed to use M twice as if they were the same, but couldn't
possibly be as you should write MG/r^2 and MG/c^2r instead of M/r^2
and 2M/r.
John Polasek
[Hammond]
Na, na, na...... Newton's equation is written as:
force/unit mass = F/m = -GM/r^2
there is ONLY ONE M in this expression for the acceleration
(force/unit mass) and that is the M causing the gravitational
field.
Note, in the formula above for the Christoffel Symbol, units
where G = c = 1 are assumed, as they are in all of advanced
physics. It's the SAME M in both formulas. G and c can be
reinserted in the formulas or not depending on your preference
for units.
========================================
SCIENTIFIC PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
=======================================
Join COSA church (Church of the Scientific Advent)
Send a blank email to
and your email address will be added to the
COSA discussion list (free, no obligation)
===========================
and please ask your news service to add:
alt.sci.relativistic-proof-of-god.moderated
===========================
.
|
|
|
|
|
| User: "brian a m stuckless" |
|
| Title: Re: Newton's from Einstein? |
17 Sep 2005 03:59:12 PM |
|
|
[GPSbob.wpd]; GPS & c^2 / f_b
The relative difference in the rate of a clock in a circular orbit of
EARth compared to a clock 'in' a Very Low Orbit (VLO) 'at' EQUiVALENT
radius as EARth (NOT "on" EARth's surface ..as some scientists say):
The GPS ATOMiC clock PREset Factor is 4.46 x 10^-10
= (Very Low Orbit period tb) / (t1 - tb)
= GPS orbital AMBiENT density / SURFACE AMBiENT density d3
= (GPS AMBiENT particle COUNT/m^3) / Loschmidt's Constant ..no
= (1.20 x 10^16 molpart/m^3) / (2.69 x 10^25 molpart/m^3)
= (v1^2 - vb^2) / 2*c^2
= (rA*gA - r3*g3) / 2*c^2
= G*M1*[{1/(n1 - 1)*rA} - {1/(n3 - 1)*r3}] / c^2
(fA - fb) / fb
= [{G*M1 / {(n1 - 1)*rA}} - {G*M1 / {(n3 - 1)*r3}}] / c^2.
1. CAViTY n=0 only & always in GR & Newton Tivity (BOTH ..no AMBiENT).
2. The GUESS Kinematic Viscosity & Diffusion factor = c^2 / fb
= G*M1*[{1/(n1 - 1)*rA} - {1/(n3 - 1)*r3}] / (fA - fb).
3. v1^2 = G*M1 / (n1 - 1)*rA = rA*gA.
4. Vb^2 = G*M1 / (n3 - 1)*r3 = r3*g3 ..Note delta ALTiTUDE, (rA - r3).
5. The CAViTY n of n1 & n3 of mass m1 includes any m1 LATTiCE VACANCY.
M1 = Mass of EARth (here).
G = Gravitational Constant.
r3 = radius of EARth or rocket VLO.
rA = radius of orbiting clock's orbit.
v1 = 2*pi*rA/t1 = orbital speed of orbiting clock.
vb = 2*pi*r3/t3 = speed of the EARth clock (ECi frame).
v1^2 = rA*g = (vescape)^2 / 2*(n - 1) = G*M1 / (n - 1)*rA
[ m1*c/h=nL/wl=fL/c=pL/h=nA/2*pi*rA=eV/h*fA*rA=pA/h*rA ].
<> >><> >><> >><> >><>
("`-/")_.-'"``-._
\. . `; -._ )-;-, `)
Yours sincerely, \ / (v_,) _ )`-.\ ``-'
```brian a m stuckless - O - _.- _..-_/ / ((.'
TiME 07:29iAMsat17sep2005; / \ ((,.-' ((,/ By: Toe.!
p.s.
(n - 1) = (mD - m1) / m1 = Fine Structure VARiABLE,
--where mD is EQUiVALENT AMBiENT DisCHARGE mass from m1
--and where (n) = mD / m1 ..provides the + or - sign.
No more is more.
*:-.,_,.-:*'``'*:-.,_,.-:*'``'*:-.,_,.-:*'``'*:-.,_,.-:*'`
____ _ _ _ _
| _ \ | | ___ _ __ | | __ | | | |
| |_) | | | / _ \ | '_ \ | |/ / | | | |
The BiG | __/ | | | (_) | | | | | | < _ |_| |_|
|_| |_| \___/ |_| |_| |_|\_\ (_) (_) (_)
_.-:*'``'*:-.,_,.-:*'``'*:-.,_,.-:*'``'*:-.,_,.-:*'``'*:-.
.
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