GR to Newton. Unable to get there from here.
Simple Simon
http://upload.wikimedia.org/math/e/d/0/ed0b180aff4406023a0549e39e6d371d.png
and
http://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d35a.png
from
http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric
and simplifying assumptions (such as d\theta =0 and d\phi = 0, and a small
mass about the size of the earth and a radial distance about the size of the
earth's radius) I would hope to be able to arrive at the "Newtonian
approximation (?)" of the acceleration due to gravity:
d^2r/d\Tau^2 = (GM) / (r^2)
So far, when I try (my derivative gets messy) I get stuck.
It would be very nice if someone could show me the way.
Thank you.
Eric Gisse
> Givenhttp://upload.wikimedia.org/math/e/d/0/ed0b180aff4406023a0549e39e6d37...
> andhttp://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d3...
> fromhttp://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_m...
>
> and simplifying assumptions (such as d\theta =0 and d\phi = 0, and a small
> mass about the size of the earth and a radial distance about the size of the
> earth's radius) I would hope to be able to arrive at the "Newtonian
> approximation (?)" of the acceleration due to gravity:
> d^2r/d\Tau^2 = (GM) / (r^2)
>
> So far, when I try (my derivative gets messy) I get stuck.
> It would be very nice if someone could show me the way.
> Thank you.
Geodesic equation.
Drop all terms higher than 1/r^2.
Tom Roberts
You must consider the equations of motion of the earth in the
gravitational field of the sun. In Newtonian gravity this is "F=mg" with
"g=-mM/r^2"; in GR this is the geodesic equation of Schw. spacetime. So
attempting to approximate the metric is not what you need to do.
I suggest looking in a good textbook on GR, such as Schutz.
Tom Roberts
Juan R.
> Given
> http://upload.wikimedia.org/math/e/d/0/
ed0b180aff4406023a0549e39e6d371d.png
> and
> http://upload.wikimedia.org/math/d/e/3/
de30517d19e3c99181aa0ccf9393d35a.png
> from
> http://en.wikipedia.org/wiki/
Schwarzschild_metric#The_Schwarzschild_metric
>
> and simplifying assumptions (such as d\theta =0 and d\phi = 0, and a
> small mass about the size of the earth and a radial distance about the
> size of the earth's radius) I would hope to be able to arrive at the
> "Newtonian approximation (?)" of the acceleration due to gravity:
> d^2r/d\Tau^2 = (GM) / (r^2)
>
> So far, when I try (my derivative gets messy) I get stuck. It would be
> very nice if someone could show me the way. Thank you.
Unfortunately, it is not possible to go from GR to Newton.
It is not possible to obtain a flat metric the right field equations and
a nonzero acceleration all at once in GR.
Textbook derivations are not acceptable because they are based in a
number of invalid steps. E.g. authors as Carroll usually falsify the weak
field geodesic equation simply substituting the result predicted by GR
a = 0
by the waited result
a = -grad(phi)
Wald textbook is more rigorous and blames the weak field 'derivation' by
this motive. However, Wald makes other mistakes regarding the field
equation and the spatial metric gamma_ij.
In rigor General Relativity does not reduces to Newtonian Gravity.
In Feynman formulation of gravity (FTG), one can obtain the field
equations, flat metric gamma_ij and nonzero acceleration. Feynman theory
is not geometrical but based in gravitational field.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Tom Roberts
> Unfortunately, it is not possible to go from GR to Newton.
This is not true. It is a simple matter to compare the orbits of planets
in GR and NG, and see they are observationally indistinguishable, except
for certain cases like the precession of the perihelion of Mercury. It
is also easy to compare the two theories' predictions for thrown
baseballs and the like, and note they are experimentally
indistinguishable for everyday situations.
[NG = Newtonian gravity]
> It is not possible to obtain a flat metric the right field equations and
> a nonzero acceleration all at once in GR.
This is true, but is unrelated to the Newtonian APPROXIMATION to GR.
> [...]
> In rigor General Relativity does not reduces to Newtonian Gravity.
The issue is not "reducing" to NG, but rather being APPROXIMATED by NG.
The former cannot be done, but the latter can -- because the predictions
for the two theories are indistinguishable in the regime where their
domains of applicability overlap.
> In Feynman formulation of gravity (FTG), one can obtain the field
> equations, flat metric gamma_ij and nonzero acceleration. Feynman theory
> is not geometrical but based in gravitational field.
But his formulation is equivalent to GR, locally (which is all that
matters here). If one can do it for FTG, one can do it for GR.
Tom Roberts
Ken S. Tucker
> Givenhttp://upload.wikimedia.org/math/e/d/0/ed0b180aff4406023a0549e39e6d37...
> andhttp://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d3...
> fromhttp://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_m...
>
> mass about the size of the earth and a radial distance about the size of the
> earth's radius) I would hope to be able to arrive at the "Newtonian
> approximation (?)" of the acceleration due to gravity:
> d^2r/d\Tau^2 = (GM) / (r^2)
>
> So far, when I try (my derivative gets messy) I get stuck.
> It would be very nice if someone could show me the way.
> Thank you.
Hi Simon.
You were kind enough to help me link to GR1916
so go to Eq.(67) therein. It's quite clear, but
if you have a problem, I'll watch this thread
and help out. The chptr it's in "Newton's Theory
as a First Approximation" is worth a read.
Regards
Ken S. Tucker
Juan R.
> Juan R. González-Álvarez wrote:
>> Unfortunately, it is not possible to go from GR to Newton.
(...)
>> It is not possible to obtain a flat metric the right field equations
>> and a nonzero acceleration all at once in GR.
(...)
>> [...]
>> In rigor General Relativity does not reduces to Newtonian Gravity.
(sniped new lie by Tom Roberts)
>> In Feynman formulation of gravity (FTG), one can obtain the field
>> equations, flat metric gamma_ij and nonzero acceleration. Feynman
>> theory is not geometrical but based in gravitational field.
>
> But his formulation is equivalent to GR, locally (which is all that
> matters here). If one can do it for FTG, one can do it for GR.
This is plain wrong. FTG is not locally equivalent to GR
Moreover, a number of experiments have been recently proposed to
experimentally differentiate FTG from GR.
http://www.canonicalscience.org/en/publicationzone/
canonicalsciencetoday/20080516.html
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Dirk Van de moortel
Y5adnUeka5F34VnV...@comcast.com
Writing R = G M/c^2, and using accent-notation for differentiation
w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
the Lagrangian (formally divice the metric by (d\Tau)^2
L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
The geodesic equations are
[ @L/dt' ]' - @L/dt = 0 [2]
[ @L/dr' ]' - @L/dr = 0 [3]
Since L has no explicit dependence on t, you have from [2]
[ @L/dt' ]' = 0
and thus
@L/@t' = k1 (some constant)
which gives with [1]
(1-R/r) t' = k1 / (2 c^2) = k2 (another constant) [4]
or
t' = k2 / (1-R/r)
Substitute this into [1] to get
c^2 = c^2 (k2)^2 / (1-R/r) - r'^2 / (1-R/r)
or
c^2 (1-R/r) = c^2 (k2)^2 - r'^2
or
r'^2 = c^2 (k2)^2 - c^2 (1-R/r)
or
r'^2 = k3 - c^2 (1-R/r) (yet another constant)
or differenting w.r.t Tau:
2 r' r" = c^2 R r' / r^2 (we have lost the constant :-) )
or simplifying:
r" = c^2 R / r^2
or, using R = G M/c^2
r'' = G M / r^2
or, using your notation
d^2r/d\Tau^2 = G M / (r^2)
Of course this *is* not Newtonian, since you have proper time Tau
and reduced circumference r. It approximates Newtonian at infinity.
Dirk Vdm
Dirk Van de moortel
[reposted with a few typos corrected]
Writing R = G M/c^2, and using accent-notation for differentiation
w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
the Lagrangian (formally divide the metric by (d\Tau)^2)
L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
The geodesic equations are
[ @L/dt' ]' - @L/dt = 0 [2]
[ @L/dr' ]' - @L/dr = 0 [3]
Since L has no explicit dependence on t, you have from [2]
[ @L/dt' ]' = 0
and thus
@L/@t' = k1 (some constant)
which gives with [1]
(1-R/r) t' = k1 / (2 c^2) = k2 (another constant) [4]
or
t' = k2 / (1-R/r)
Substitute this into [1] to get
c^2 = c^2 (k2)^2 / (1-R/r) - r'^2 / (1-R/r)
or
c^2 (1-R/r) = c^2 (k2)^2 - r'^2
or
r'^2 = c^2 (k2)^2 - c^2 (1-R/r)
or
r'^2 = k3 - c^2 (1-R/r) (yet another constant)
or differenting w.r.t Tau:
2 r' r" = - c^2 R r' / r^2 (we have lost the constant :-) )
or simplifying:
r" = - c^2 R / r^2
or, using R = G M/c^2:
r'' = - G M / r^2
or, using your notation:
d^2r/d\Tau^2 = - G M / (r^2)
Of course this is not Newtonian, since you have proper time Tau
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Simple Simon wrote on Sun, 07 Sep 2008 18:16:01 -0800:
>
> > Given
> >http://upload.wikimedia.org/math/e/d/0/
>
> ed0b180aff4406023a0549e39e6d371d.png> and
> >http://upload.wikimedia.org/math/d/e/3/
>
> de30517d19e3c99181aa0ccf9393d35a.png> from
> >http://en.wikipedia.org/wiki/
>
> Schwarzschild_metric#The_Schwarzschild_metric
>
>
>
> > and simplifying assumptions (such as d\theta =0 and d\phi = 0, and a
> > small mass about the size of the earth and a radial distance about the
> > size of the earth's radius) I would hope to be able to arrive at the
> > "Newtonian approximation (?)" of the acceleration due to gravity:
> > d^2r/d\Tau^2 = (GM) / (r^2)
>
> > So far, when I try (my derivative gets messy) I get stuck. It would be
> > very nice if someone could show me the way. Thank you.
>
> Unfortunately, it is not possible to go from GR to Newton.
Oh my god will you please stop lying about about this?
>
> It is not possible to obtain a flat metric the right field equations and
> a nonzero acceleration all at once in GR.
Of course it isn't, Juan. This is stupid and wrong for two reasons:
1) Nobody ever claimed the flat metric has gravitation. That's one of
your misunderstandings that nobody can cure.
2) There is zero acceleration in GR - objects move on _geodeics_.
Remember what those are?
>
> Textbook derivations are not acceptable because they are based in a
> number of invalid steps. E.g. authors as Carroll usually falsify the weak
> field geodesic equation simply substituting the result predicted by GR
>
> a = 0
>
> by the waited result
>
> a = -grad(phi)
...and when asked to show where the claimed "substitution" occures,
you clam up or blow smoke. You are yet to actually explain what's
wrong with the derivation used in every textbook and introduction to
weak field limits I have ever seen.
>
> Wald textbook is more rigorous and blames the weak field 'derivation' by
> this motive. However, Wald makes other mistakes regarding the field
> equation and the spatial metric gamma_ij.
He makes *other* mistakes now? Well when you are going to spew
nonsense, might as well aim high.
>
> In rigor General Relativity does not reduces to Newtonian Gravity.
Hey weren't you supposed to have a paper published on this? Whatever
happened with that - did people notice it was full of shit and forward
it to the circular file where it belongs?
>
> In Feynman formulation of gravity (FTG), one can obtain the field
> equations, flat metric gamma_ij and nonzero acceleration. Feynman theory
> is not geometrical but based in gravitational field.
>
> --http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
....and how is FTG relevant? Hint: IT ISN'T.
Every time you come to propagate this nonsense, you *always* have to
drop in irrelevant shit about other theories of gravitation. Why?
Juan R.
I am not going to waste any time discussing basic stuff with a well-known
crackpot as you:
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
ii.html
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
[...whining...]
What's the matter Juan, your paper get rejected again?
Is that why you are back to lying about GR again on USENET?
Dirk Van de moortel
8mfxk.204112$_M4.1...@newsfe18.ams2
typos:
[ @L/@t' ]' - @L/@t = 0 [2]
[ @L/@r' ]' - @L/@r = 0 [3]
[ @L/@t' ]' = 0
> and thus
> @L/@t' = k1 (some constant)
> which gives with [1]
> (1-R/r) t' = k1 / (2 c^2) = k2 (another constant) [4]
> or
> t' = k2 / (1-R/r)
>
> Substitute this into [1] to get
> c^2 = c^2 (k2)^2 / (1-R/r) - r'^2 / (1-R/r)
> or
> c^2 (1-R/r) = c^2 (k2)^2 - r'^2
> or
> r'^2 = c^2 (k2)^2 - c^2 (1-R/r)
> or
> r'^2 = k3 - c^2 (1-R/r) (yet another constant)
> or differenting w.r.t Tau:
> 2 r' r" = - c^2 R r' / r^2 (we have lost the constant :-) )
> or simplifying:
> r" = - c^2 R / r^2
> or, using R = G M/c^2:
> r'' = - G M / r^2
> or, using your notation:
> d^2r/d\Tau^2 = - G M / (r^2)
>
> Of course this is not Newtonian, since you have proper time Tau
> and reduced circumference r. It approximates Newtonian at infinity.
And this should be more like...
Of course this is not Newtonian, since you have proper time Tau,
reduced circumference r and arbitrary speeds.
At infinity and for low speeds, Newtonian gravity approximates it.
Dirk Vdm
Ken S. Tucker
On Sep 9, 4:52 am, "Dirk Van de moortel"
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Dirk Van de moortel <dirkvandemoor...@nospAm.hotmail.com> wrote in message
> 8mfxk.204112$_M4.155...@newsfe18.ams2
>
>
>
> > Simple Simon <pi.r.cubed-nos...@gmail.com> wrote in message
> > Y5adnUeka5F34VnVnZ2dnUVZ_sbin...@comcast.com
> >> Given
> >>http://upload.wikimedia.org/math/e/d/0/ed0b180aff4406023a0549e39e6d37...
> >> and
> >>http://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d3...
> >> from
> >>http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_m...
With all that effort, would you care to provide
the "L" you're using above?
I think your treatment is too complex especially
for the OP asking a straighforward question about
the geodesic, but we're prepared to listen to your
reasoning.
Regards
Ken S. Tucker
Eric Gisse
See that equation called [1], with the L = in it? Try reading that.
Dirk derives L from the metric itself, which is described in most
proper GR textbooks. Which you have never read.
> I think your treatment is too complex especially
> for the OP asking a straighforward question about
> the geodesic, but we're prepared to listen to your
> reasoning.
> Regards
> Ken S. Tucker
No, stupid. That's exactly what the OP was asking for.
Ken S. Tucker
To add.
As a high school student I found GR1916 Eq.(67)
quite clear. AE didn't use Lagrange therein at
all, and Weinberg mentions it once in "Grav&Cosmo",
it's not important in GR. If anything, focus your
effort on Hamilton's Principle, but you'll need to
get up to speed on Relative Tensors.
Regards
Ken S. Tucker
Dirk Van de moortel
93c6d17c-f525-4862...@1g2000pre.googlegroups.com
What effort?
> would you care to provide
> the "L" you're using above?
In short, with the same accent notation X' for dX / d\Tau:
Total proper time = integral sqrt( g_mn dx^m dx^n )
= integral sqrt( g_mn x'^m x'^n ) d\Tau
Motion in free fall extremizes total proper time, so
g_mn x'^m x'^n is a Lagrangian and satifies the Euler
equations, [2] and [3] being two of them.
For radial motion in free fall
d\theta =0 and d\phi = 0
so
theta' = phi' = 0
so from the metric:
L( x^m, x'^m ) = g_mn x'^m x'^n
= (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) = c^2
> I think your treatment is too complex especially
> for the OP asking a straighforward question about
> the geodesic, but we're prepared to listen to your
> reasoning.
I gave a straighforward answer about the geodesic.
This stuff should appear in the first chapters of introductions
to GR, and in abundance on the net (for instance on page 10
of http://physics.gmu.edu/~joe/PHYS428/Topic10.pdf )
If it is too compex for you, then that's just fine with me.
Dirk Vdm
Ken S. Tucker
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote in message
> 93c6d17c-f525-4862-8ba4-b883a45f4...@1g2000pre.googlegroups.com
Ok, you're using AE's GR1916 Eq.(20) to derive
the geodesic, (apart from minor notational's
you're in the ball-park.
> > I think your treatment is too complex especially
> > for the OP asking a straighforward question about
> > the geodesic, but we're prepared to listen to your
> > reasoning.
>
> I gave a straighforward answer about the geodesic.
> This stuff should appear in the first chapters of introductions
> to GR, and in abundance on the net (for instance on page 10
> ofhttp://physics.gmu.edu/~joe/PHYS428/Topic10.pdf)
>
> If it is too compex for you, then that's just fine with me.
The OP didn't ask for the means to derive the
geodesic equation, but instead, how to apply
it, read his 1st post, he's quite clear and I
think AE does a good job of that in GR 1916
Eq.(67).
Btw, I think your approach is obsolete, have a
look at something more modern,
U^u;v U^w == DU^u/ds = 0
is much simpler and physically intuitive.
Regards
Ken S. Tucker
Dirk Van de moortel
> On Sep 9, 12:06 pm, "Dirk Van de moortel"
[snip]
>> I gave a straighforward answer about the geodesic.
>> This stuff should appear in the first chapters of introductions
>> to GR, and in abundance on the net (for instance on page 10
>> ofhttp://physics.gmu.edu/~joe/PHYS428/Topic10.pdf)
>>
>> If it is too compex for you, then that's just fine with me.
>
> The OP didn't ask for the means to derive the
> geodesic equation, but instead, how to apply
> it,
Go imitate your roof and take a leak.
Dirk Vdm
Koobee Wublee
> [reposted with a few typos corrected]
It does not matter how many times you have corrected yourself, it is
still wrong. <shrug>
> Writing R = G M/c^2, and using accent-notation for differentiation
> w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
> the Lagrangian (formally divide the metric by (d\Tau)^2)
> L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
So, you have been reading yours-truly’s work but cannot even copy down
what I wrote correctly. The Lagrangian is
** L = c^2 (1 – 2 R / r) (dt/ds)^2 – (dr/ds)^2 / (1 – 2 R / r) - ...
= 1
Where
** ds^2 = c^2 (1 – 2 R / r) dt^2 – dr^2 / (1 – 2 R / r)
** ds^2 = Your c^2 dTau^2
** dt/ds = Your t’
** dr/ds = Your r’
> The geodesic equations are
> [ @L/dt' ]' - @L/dt = 0 [2]
> [ @L/dr' ]' - @L/dr = 0 [3]
>
> Since L has no explicit dependence on t, you have from [2]
> [ @L/dt' ]' = 0
> and thus
> @L/@t' = k1 (some constant)
> which gives with [1]
> (1-R/r) t' = k1 / (2 c^2) = k2 (another constant) [4]
> or
> t' = k2 / (1-R/r)
dt/ds = k2 / (1 – 2 R / r)
> Substitute this into [1] to get
> c^2 = c^2 (k2)^2 / (1-R/r) - r'^2 / (1-R/r)
> or
> c^2 (1-R/r) = c^2 (k2)^2 - r'^2
> or
> r'^2 = c^2 (k2)^2 - c^2 (1-R/r)
> or
> r'^2 = k3 - c^2 (1-R/r) (yet another constant)
(dr/ds)^2 = k3 + 2 R / r
> or differenting w.r.t Tau:
> 2 r' r" = - c^2 R r' / r^2 (we have lost the constant :-) )
> or simplifying:
> r" = - c^2 R / r^2
2 dr/ds d^2r/ds^2 = - 2 R c^2 dr/ds / r^2
> or, using R = G M/c^2:
> r'' = - G M / r^2
> or, using your notation:
> d^2r/d\Tau^2 = - G M / (r^2)
d^2r/ds^2 = - R / r^2
> Of course this is not Newtonian, since you have proper time Tau
> and reduced circumference r. It approximates Newtonian at infinity.
Of course, the geodesic model where each path follows the least amount
of accumulated spacetime is absurd. It does not allow photons to
propagate.
Now, show us how you derive the above equation using the other Euler-
Lagrange equation [3]. Better yet, show us the Lagrangian where the
geodesic path follows the one with the least amount of time according
to Fermat’s principle. Then, derive Newton’s law of gravitation based
on that. See if you see the surprise.
Ken S. Tucker
<juanREM...@canonicalscience.com> wrote:
> Eric Gisse wrote on Mon, 08 Sep 2008 12:52:44 -0700:
> I am not going to waste any time discussing basic stuff with a well-known
> crackpot as you:
I nailed Gisse to the cross awhile back.
The twink did a particularily good post, that
was far beyond his/her level of comprehension
and expertise so I suspected plagurism, sure
enough it was.
So I started asking a few questions that only
an expert level poster who had truly formulated
the post could answer.
Well it didn't take look before the twink moaned
like a sissy, with the usual claim that the
questions were unfair. Meanwhile, he didn't know
that I knew his post was plagurized ;-).
It's very hard to fool the experts, one or two
probing questions exposes the twinks. That's
the system I use, like shootin' fish in a barrel.
That's part of how I qualify posters.
Regards
Ken S. Tucker
Dirk Van de moortel
aadd728f-cf8e-4b98...@t1g2000pra.googlegroups.com
[snip]
Just what I was waiting for.
I had this message ready for when you would show up.
Another predictable retired aerospace engineer, this time one
with A Very Own Private Lagrangian:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PrivateLagrangian.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewLagrangian.html
Dirk Vdm
Ken S. Tucker
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote in message
>
> > On Sep 9, 12:06 pm, "Dirk Van de moortel"
>
> [snip]
>
> >> I gave a straighforward answer about the geodesic.
> >> This stuff should appear in the first chapters of introductions
> >> to GR, and in abundance on the net (for instance on page 10
> >> ofhttp://physics.gmu.edu/~joe/PHYS428/Topic10.pdf)
>
> >> If it is too compex for you, then that's just fine with me.
>
> > The OP didn't ask for the means to derive the
> > geodesic equation, but instead, how to apply
> > it,
>
> Go imitate your roof and take a leak.
Now your talkin' my language!
The "geodesic" is solable when the piss
is NOT in "freefall".
> Dirk Vdm
Regards
Ken S. Tucker
PS: I'm high on Codeine, so I'm a bit dumb.
Dirk Van de moortel
You are low on dopamine.
Dirk Vdm.
Eric Gisse
> On Sep 9, 1:21 am, "Juan R." González-Álvarez
>
> <juanREM...@canonicalscience.com> wrote:
> > Eric Gisse wrote on Mon, 08 Sep 2008 12:52:44 -0700:
> > I am not going to waste any time discussing basic stuff with a well-known
> > crackpot as you:
>
> I nailed Gisse to the cross awhile back.
> The twink did a particularily good post, that
> was far beyond his/her level of comprehension
> and expertise so I suspected plagurism, sure
> enough it was.
Really Ken?
Since I apparently plagarized something, you should be able to not
only document what I said but what I apparently copied. Let's see if
you can provide!
Ken S. Tucker
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Ken S. Tucker <dynam...@vianet.on.ca> wrote in message
Don't know what that is, but's it my fault,
I rearranged the office-lab last night for a
new thingy project and seriously strained a
muscle, woke up in pain, guess I pushed a bit
hard. I hate drugs but the pain was causing me
to sweat. If it doesn't quit in a few days, I'm
off to a doctor, haven't been to one in 5 years.
I'll keep you all posted.
The down-side of codeine is pooping. Last time
(years ago) I bombed up on codeine, I didn't
poop for week, then when I did, it felt like I
was giving birth.
I'm not crazy about using Moraphine because it
provides an artificial sense of well-being, maybe
that's the dopemine you (Dirk) referred to, but
it's necessary for extreme pain, like burns.
I meant what I said about trickle interpretation
of the minimization of water paths. The fella
who stressed that to me was a colleague and a
world famous prof. and used integration between
points. OTOH, I preferred the absolute derivative
DU^u/ds=0 , aka intrinsic derivative, but I'm open
to suggestions where the two methods might unite
quantum theory and continuum theory in one math.
That's quite interesting, apparently the use of
integration is preferred in QT, but imv the
derivative is more powerful in classical GR.
Is that math or philosophy?
Regards
Ken S. Tucker
Koobee Wublee
> Koobee Wublee <koobee.wub...@gmail.com> wrote in message
> [snip]
You f*cked up. You don’t know what you are doing. <shrug>
You have shown using the following moortel’s fantasy metric that you
are able to fudge the Newtonian law of gravity.
ds^2 = c^2 (1 – U) dt^2 – dr^2 / (1 – U) – r^2 dO^2
Where
** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
For your reference, the Schwarzschild metric that yields the Newtonian
law of gravity is the following. Pay attention to (2 U).
ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2
You are not in the same league as the more prominent Einstein
Dingleberries in fudging the mathematics involved. They are true
‘mathemaGicians’ while you are just a spermless amateur craving for
more sperm. Even the multi-year super-senior would not come to your
rescue. Ahahahaha...
As the janitor from Cornell would say “you ought to be grateful to
[blah blah blah] for pointing out and correcting your mistakes”, yet
you are just behaving childishly by emphasizing your gross blunder.
<shrug> Also, wasting almost 20 years of your time in this newsgroup,
you have not gone beyond SR. You ought to be ashamed of yourself if
you have any bit of self-consciousness. <shrug>
Eric Gisse
> On Sep 9, 1:21 pm, "Dirk Van de moortel" wrote:
>
> > Koobee Wublee <koobee.wub...@gmail.com> wrote in message
> > [snip]
>
> You f*cked up. You don’t know what you are doing. <shrug>
>
> You have shown using the following moortel’s fantasy metric that you
> are able to fudge the Newtonian law of gravity.
>
> ds^2 = c^2 (1 – U) dt^2 – dr^2 / (1 – U) – r^2 dO^2
>
> Where
>
> ** U = G M / c^2 / r
> ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
>
> For your reference, the Schwarzschild metric that yields the Newtonian
> law of gravity is the following. Pay attention to (2 U).
>
> ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2
That isn't Schwarzschild.
Koobee Wublee
> On Sep 9, 8:54 pm, Koobee Wublee wrote:
> > You have shown using the following moortel’s fantasy metric that you
> > are able to fudge the Newtonian law of gravity.
>
> > ds^2 = c^2 (1 – U) dt^2 – dr^2 / (1 – U) – r^2 dO^2
>
> > Where
>
> > ** U = G M / c^2 / r
> > ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
>
> > For your reference, the Schwarzschild metric that yields the Newtonian
> > law of gravity is the following. Pay attention to (2 U).
>
> > ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2
>
> That isn't Schwarzschild.
You are very vague in your comment above. It is the Schwarzschild
metric but was never discovered by Schwarzschild. Hilbert was the one
who is credited for discovering the Schwarzschild metric. I suppose
that is what you meant. <shrug>
Dirk Van de moortel
> Thanks to all, I look forward to following your advice.
>
> Dirk Van de moortel:
> Thank you especially for your presentation.
>
> [ @L/@t' ]'
> looks to me like
> "The derivative with respect to tau, of the Integral part of the partial
> derivative of L with respect to the derivative of t with respect to tau"
> I expect to learn to interpret this.
No Intergral part here :-)
"The derivative with respect to tau, of the partial derivative of L
with respect to the derivative of t with respect to tau"
I could have written it as
d/d\Tau ( @L/@t' )
but I prefer the accent.
Yes, this is all standard Euler-Lagrange lingo.
It's a good idea to check up on this in the context of classical
mechanics first.
To get some idea: http://en.wikipedia.org/wiki/Euler-Lagrange
L is a function of the coordinates and their derivatives, written as
L( x^m, x'^m )
The derivatives are usually written in dot-notation, which is not
possible in ASCII, so I use accent-notation.
>
> I seem to have lost a factor of 2:
> "
> 2 r' r" = c^2 R r' / r^2 (we have lost the constant )
> or simplifying:
> r" = c^2 R / r^2
> "
Ah yes, sorry, another pair of typos:
2 r' r" = - c^2 R r' / r^2 (we have lost the constant :-) )
or simplifying:
r" = - 1/2 c^2 R / r^2 (typo 1: lost factor 1/2)
or, using R = 2 G M/c^2: (typo 2: lost factor 2)
r'' = - G M / r^2
or, using your notation:
d^2r/d\Tau^2 = - G M / (r^2)
I will repost the entire thing without the typos :-)
Dirk Vdm
Dirk Van de moortel
> Given
> http://upload.wikimedia.org/math/e/d/0/ed0b180aff4406023a0549e39e6d371d.png
> and
> http://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d35a.png
> from
> http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric
>
> mass about the size of the earth and a radial distance about the size of the
> earth's radius) I would hope to be able to arrive at the "Newtonian
> approximation (?)" of the acceleration due to gravity:
> d^2r/d\Tau^2 = (GM) / (r^2)
>
> So far, when I try (my derivative gets messy) I get stuck.
> It would be very nice if someone could show me the way.
> Thank you.
[reposted with a few more typos corrected and added a note]
Writing R = 2 G M/c^2, and using accent-notation for differentiation
w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
the Lagrangian (formally divide the metric by (d\Tau)^2)
L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
The geodesic equations for t and r are
[ @L/@t' ]' - @L/@t = 0 [2]
[ @L/@r' ]' - @L/@r = 0 [3]
| Note:
| Total proper time = integral sqrt( g_mn dx^m dx^n )
| = integral sqrt( g_mn x'^m x'^n ) d\Tau
| Motion in free fall extremizes total proper time, so
| g_mn x'^m x'^n is a Lagrangian and satifies the Euler
| equations, [2] and [3] being two of them.
|
| For radial motion in free fall
| d\theta =0 and d\phi = 0
| so
| theta' = phi' = 0
| so from the metric:
| L( x^m, x'^m ) = g_mn x'^m x'^n
| = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) = c^2
Since L has no explicit dependence on t, you have from [2]
[ @L/@t' ]' = 0
and thus
@L/@t' = k1 (some constant)
which gives with [1]
(1-R/r) t' = k1 / (2 c^2) = k2 (another constant)
or
t' = k2 / (1-R/r)
Substitute this into [1] to get
c^2 = c^2 (k2)^2 / (1-R/r) - r'^2 / (1-R/r)
or
c^2 (1-R/r) = c^2 (k2)^2 - r'^2
or
r'^2 = c^2 (k2)^2 - c^2 (1-R/r)
or
r'^2 = k3 - c^2 (1-R/r) (yet another constant)
or differenting w.r.t Tau:
2 r' r" = - c^2 R r' / r^2 (we have lost the constant :-) )
or simplifying:
r" = - 1/2 c^2 R / r^2
or, using R = 2 G M /c^2:
r'' = - G M / r^2
or, using your notation:
d^2r/d\Tau^2 = - G M / (r^2)
Of course this is not Newtonian, since you have proper time Tau,
Daryl McCullough
>Writing R = G M/c^2, and using accent-notation for differentiation
>w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
>the Lagrangian (formally divide the metric by (d\Tau)^2)
> L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
I think that you are missing a factor of 2. The correct
expression for dTau^2 is
dTau^2 = (1-2GM/(c^2 r)) dt^2
- 1/(1-2GM/(c^2 r)) dr^2/c^2
- r^2 dTheta^2/c^2
- r^2 sin(Theta)^2 dPhi^2/c^2
--
Daryl McCullough
Ithaca, NY
Dirk Van de moortel
ga8i3...@drn.newsguy.com
> Dirk Van de moortel says...
>
>> Writing R = G M/c^2, and using accent-notation for differentiation
>> w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
>> the Lagrangian (formally divide the metric by (d\Tau)^2)
>> L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
>
> I think that you are missing a factor of 2.
Yes, in the article the OP pointed to, R is defined as
R = r_s = 2 G M/c^2
http://en.wikipedia.org/wiki/Schwarzschild_metric
My typo was higher up :-)
> The correct
> expression for dTau^2 is
>
> dTau^2 = (1-2GM/(c^2 r)) dt^2
> - 1/(1-2GM/(c^2 r)) dr^2/c^2
> - r^2 dTheta^2/c^2
> - r^2 sin(Theta)^2 dPhi^2/c^2
Yes, but in the OP's setup d\Theta = d\Phi = 0 and as I said,
R = 2 G M/c^2, following the article's defs.
I have it corrected in my last message.
Dirk Vdm
Ken S. Tucker
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> Daryl McCullough <stevendaryl3...@yahoo.com> wrote in message
>
> ga8i3102...@drn.newsguy.com
If anyone has Weinberg's "Grav&Cosmo", the
Lagrangian is well explained beginning on
pg.220. An interesting equation is simply,
dtau/dt = 1 - L , Eq.(9.2.2),
which may be a bit more reliable than Wiki.
Regards
Ken S. Tucker
Ken S. Tucker
This could be fun.
Let me call dtau = ds, then with a bit of
calculus find, (a good approximation)...
(using Eq.(9.2.2) above),
d^2(t) = dL ds , d^2(s)= -dL dt
and see how the time paradox developes using
the Lagrangian "L", in those.
The "ds" can be thought of as the uniform rate
at which a clocks ticks in CS k and likewise
for dt in CS K, and the d^2(t) and d^2(s) are
the accelerations of those clock *rates* as
"defined" by the change in the Lagrangian "dL".
Correct me if I'm wrong: In Newton Theory "L"
is constant so that dL=0 and dt^2 =0 because
both dt and ds are sync'd constant...tick, tick.
If I understand correctly, there is a subtle
variation in the Lagrange "L" in transforming
from Newton's theory to GR, wherein GR, L is
not constant.
Is that right?
Regards
Ken S. Tucker
Eric Gisse
[snip]
Hey asshole, how DARE you claim plagarism on my part?! You are such a
pussy that you will not even substantiate your claims!
Koobee Wublee
> Writing R = 2 G M/c^2, [...]
I have already corrected your gross fumble with the following post.
You should have known better after spending almost 20 years in this
newsgroup, and there is also no need for you to repeat what I said by
inappropriately glorifying yourself.
http://groups.google.com/group/sci.physics.relativity/msg/f547abcc68207f61?hl=en
If you had listened to Koobee Wublee in the first place, you will have
understood the issue then and corrected. <shrug>
Dirk Van de moortel
> On Sep 10, 12:58 am, "Dirk Van de moortel" wrote:
>
>> Writing R = 2 G M/c^2, [...]
>
> I have already corrected your gross fumble
http://upload.wikimedia.org/math/d/e/3/de30517d19e3c99181aa0ccf9393d35a.png
http://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric
Dirk Vdm
Tom Roberts
> (sniped new lie by Tom Roberts)
If what I wrote is really a lie, you would able to display it. Please do
so. That is, show where I "lied" when I said:
> It is a simple matter to compare the orbits of planets in GR and NG,
> and see they are observationally indistinguishable, except for
> certain cases like the precession of the perihelion of Mercury. It is
> also easy to compare the two theories' predictions for thrown
> baseballs and the like, and note they are experimentally
> indistinguishable for everyday situations.
This is a clear refutation of your claim that GR is incompatible with
Newtonian gravitation (NG) (or that GR "has no Newtonian limit").
Note that I agree that "GR does not reduce to NG", but also
note that this is IRRELEVANT. What matters is that the
predictions of GR are experimentally indistinguishable from
the predictions of NG in their common domain of applicability.
That is, the experiments that established NG do not refute GR.
IOW: the relationship between GR and NG is APPROXIMATION,
not "reduction" (i.e. a series of theorems leading from one
to the other).
> Moreover, a number of experiments have been recently proposed to
> experimentally differentiate FTG from GR.
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/20080516.html
I could find no evidence of such proposed experiments there. Please be
more specific.
Tom Roberts
Eric Gisse
> Juan R. González-Álvarez wrote:
> > (sniped new lie by Tom Roberts)
>
> If what I wrote is really a lie, you would able to display it. Please do
> so. That is, show where I "lied" when I said:
>
> > It is a simple matter to compare the orbits of planets in GR and NG,
> > and see they are observationally indistinguishable, except for
> > certain cases like the precession of the perihelion of Mercury. It is
> > also easy to compare the two theories' predictions for thrown
> > baseballs and the like, and note they are experimentally
> > indistinguishable for everyday situations.
>
> This is a clear refutation of your claim that GR is incompatible with
> Newtonian gravitation (NG) (or that GR "has no Newtonian limit").
It gets better, Tom. He has _explicitly_ argued in the past that
Schwarzschild under specific limiting conditions yields Newtonian
gravitation. Which makes his whining irrelevant.
Personally I'm getting bored with his lying about the subject. Every
time this comes up, he slanders Wald & Carroll, insults me, and
misquotes people he has talked to in the past. Read how he argues that
GR has no Newtonian limit because he had it explained to him that
there is no gravitation in flat space.
[...]
Juan R.
> It gets better, Tom. He has _explicitly_ argued in the past that
> Schwarzschild under specific limiting conditions yields Newtonian
> gravitation.
As usual you cannot read. It is true that in the _past_ I had a beautiful
discussion with Steve Carlip (and others) in this newsgroup and in s.p.r.
about Newtonian limits.
Then I worked with convention x^0 = ct and Carlip with x^0 = t both
obtained Newton equation of motion. Carlip explicitly showed how my
approach gave the same results than him, after a 'renormalization'. Then
I agreed with Carlip but after further research I found mistakes in
*both* works.
Of course, I am not going to waste time explaining techical details to
crackpots like you, who does not know even the most basic stuff:
http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-fauna-
ii.html
(sniped a lie)
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
(...)
>
> Note that I agree that "GR does not reduce to NG"
>
But you are nobody. Students, other scientists and general public still
may read in GR textbooks, papers, and encyclopedias that *wrong*
statement that "GR reduces to NG". For instance,
http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation
(\blockquote
General relativity reduces to Newtonian gravity in the limit of small
potential and low velocities
)
And Wald 1984 in its page 78 states:
(\blockquote
Above, we showed that general relativity reduces to Newtonian gravity
)
> What matters is that the predictions of GR are
> experimentally indistinguishable from the predictions of NG
Exactly, and they are *not*.
It is taking me too many time to convince a stupid referee [1]. Each time
I caught it in some flagrant mistake, he changes his review and attacks
my paper with some other new stuff, which then I may show wrong...
Fortunately, I already convinced him of three main points. I copy him:
(\blockquote
Strictly speaking, Newton's theory is not contained in GR, and there
is no reason why it should be.
)
He is right and contrast with Carlip, Hillman [2], Carroll, Wald...
incorrect claims.
(\blockquote
One is a metric theory, the other is not.
)
Again right, I am really surprised each time I find other General
Relativists confounding metric with DPI theories... Carlip is a notorious
case.
My works shows, in a rigorous mathematical sense, that no metric theory
of gravity can reproduce Newtonian results in the *complete* space
dynamics (i.e. for arbitrary values of parameters).
The anonymous referee, a recognized expertise in numerical gravity, also
wrote:
(\blockquote
If the author wishes to check this using numbers instead of
limits, he could simply program a computer with the Schwarzschild
metric and the geodesic equation, and integrate numerically the motion
of tests particles for small values of M/R. He will then find that
for small enough M/R the trajectories become identical to Newtonian
trajectories to extremely high numerical accuracy (in fact, as high as
one desires for small enough M/R).
)
Right! There is a difference [3] between trajectories which is of order
(M/R). It is also true that in that *specific* case, Schwarzschild
metric, GR trajectories are practically 'indistinguishable' [4] from NG
ones for low values of (M/R).
But, and this is the point where the referee *fails*, the difference
between orbits is very important in many-body situations, which are
characteristically chaotic.
In this more general case, differences of order (M/R), no matter how
small, are *exponentially* amplified (the unitary propagator of the
dynamics is e^Lt) with the result that GR solution turns unstable and
does not correspond to the NG result.
GR gives the wrong multiparticle correlations for a generic many-body
state [5].
In the basis of that result, it is not strange that worldwide expertises
on relativistic chaos as Prof. W. C. Schieve
http://order.ph.utexas.edu/research/glimpse.html
have *abandoned* GR for the difficult task of developing a relativistic
many body dynamics:
http://order.ph.utexas.edu/mtrump/manybody/
See some equations of motion for relativistic gravity in
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-
limitations_20.html
(sniped another lie)
NOTES
[1] Of a top journal on General Relativity. Of course, that
referee is doing an effort to withdraw a paper that shows that
sophisticated General Relativity cannot reproduce some Newtonian
results, and that a more general theory of gravity is not
geometric.
[2] Once Chris Hillman authored a webpage titled "Some Scientifically
Inaccurate Claims Concerning Cosmology and Relativity" was full of
false statements. He deleted the site
http://math.ucr.edu/home/baez/relativity.html
[3] Referee obtained that using numerical methods. I got using a more
general mathematical study of a perturbative expansion of the
geodesic equation.
[4] There is still a difference regarding cosmological boundaries which
is addressed in a specific section. The GR problem of unphysical
island universe boundaries is *absent* in NG.
[5] As showed in another paper GR is built over a no-correlation
approximation. GR is obtained as special case from a more general
theory of gravity which is not geometrical.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> Eric Gisse wrote on Fri, 12 Sep 2008 17:00:56 -0700:
(...)
> As usual you cannot read. It is true that in the _past_ I had a
> beautiful discussion with Steve Carlip (and others) in this newsgroup
> and in s.p.r. about Newtonian limits.
>
> Then I worked with convention x^0 = ct and Carlip with x^0 = t both
> obtained Newton equation of motion. Carlip explicitly showed how my
> approach gave the same results than him, after a 'renormalization'. Then
> I agreed with Carlip but after further research I found mistakes in
> *both* works.
To be more clear and precise *then* (i.e. in the past) both Carlip and me
got the equation of motion (or we believed that _then_) applying the
nonrelativistic limit
c --> infinity
in the geodesic equation of motion.
But, in recent times, I developed a geometrical check (for checking
metric theories of gravity) and when I applied it to _past_ derivations I
found a subtle mathematical mistake that both Carlip and me did _then_.
After correcting the derivations, eliminating that mistake, I found _now_
that the GR geodesic equation of motion does *not* reduce to
a = - \grad \phi
in the limit c --> infinity.
> Of course, I am not going to waste time explaining techical details to
> crackpots like you, who does not know even the most basic stuff:
>
> http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
> fauna-ii.html
>
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Eric Gisse wrote on Fri, 12 Sep 2008 17:00:56 -0700:
>
> > It gets better, Tom. He has _explicitly_ argued in the past that
> > Schwarzschild under specific limiting conditions yields Newtonian
> > gravitation.
>
> As usual you cannot read. It is true that in the _past_ I had a beautiful
> discussion with Steve Carlip (and others) in this newsgroup and in s.p.r.
> about Newtonian limits.
Did I SAY Steve Carlip?
http://groups.google.com/group/sci.physics.relativity/msg/2bdb5a10ae6616a8?dmode=source
....isn't it interesting that you still are unpublished and have
returned to defending your stupidities on USENET?
>
> Then I worked with convention x^0 = ct and Carlip with x^0 = t both
> obtained Newton equation of motion. Carlip explicitly showed how my
> approach gave the same results than him, after a 'renormalization'. Then
> I agreed with Carlip but after further research I found mistakes in
> *both* works.
Yet again you make claims and do not document them.
>
> Of course, I am not going to waste time explaining techical details to
> crackpots like you, who does not know even the most basic stuff:
>
> http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-f...
> ii.html
>
> (sniped a lie)
>
> --http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
You don't waste your time *ever* explaining technical details. Even
before I gave you the excuse you needed to treat me like an idiot, you
refused to discuss the technical details of your work. You frequently
referred to private correpondence and unpublished and private works to
buttress your positions and here we are again with you holding back
details.
Why don't you self-publish your monograph, hm? That way the work can
be evaluated on its' merits and you don't have to be bitchy and whiny
about folks like me working on supposition and guesswork [not that you
give us anything else to use] as has been done since
June...July...August...September...
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> "Juan R." González-Álvarez wrote on Sat, 13 Sep 2008 10:11:20 +0200:
>
> > Eric Gisse wrote on Fri, 12 Sep 2008 17:00:56 -0700:
>
> (...)
>
> > As usual you cannot read. It is true that in the _past_ I had a
> > beautiful discussion with Steve Carlip (and others) in this newsgroup
> > and in s.p.r. about Newtonian limits.
>
> > Then I worked with convention x^0 = ct and Carlip with x^0 = t both
> > obtained Newton equation of motion. Carlip explicitly showed how my
> > approach gave the same results than him, after a 'renormalization'. Then
> > I agreed with Carlip but after further research I found mistakes in
> > *both* works.
>
> To be more clear and precise *then* (i.e. in the past) both Carlip and me
> got the equation of motion (or we believed that _then_) applying the
> nonrelativistic limit
>
> c --> infinity
>
> in the geodesic equation of motion.
>
> But, in recent times, I developed a geometrical check (for checking
> metric theories of gravity) and when I applied it to _past_ derivations I
> found a subtle mathematical mistake that both Carlip and me did _then_.
>
> After correcting the derivations, eliminating that mistake, I found _now_
> that the GR geodesic equation of motion does *not* reduce to
>
> a = - \grad \phi
>
> in the limit c --> infinity.
Who cares? The limit c---> \infty is not the Newtonian limit.
Juan R.
(snip new nonsense by crackpot Eric Gisse)
>
>> > Of course, I am not going to waste time explaining techical details
>> > to crackpots like you, who does not know even the most basic stuff:
>>
>> >http://canonicalscience.blogspot.com/2008/08/some-samples-of-usenet-
>> > fauna-ii.html
>>
>> --http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
> On Sep 13, 12:11 am, "Juan R." González-Álvarez
> <juanREM...@canonicalscience.com> wrote:
>> Eric Gisse wrote on Fri, 12 Sep 2008 17:00:56 -0700:
>>
>> > It gets better, Tom. He has _explicitly_ argued in the past that
>> > Schwarzschild under specific limiting conditions yields Newtonian
>> > gravitation.
>>
>> As usual you cannot read. It is true that in the _past_ I had a
>> beautiful discussion with Steve Carlip (and others) in this newsgroup
>> and in s.p.r. about Newtonian limits.
>
> Did I SAY Steve Carlip?
>
> http://groups.google.com/group/sci.physics.relativity/
msg/2bdb5a10ae6616a8?dmode=source
>
Which proves your previous lie.
(...)
> You don't waste your time *ever* explaining technical details.
Correction, I don't waste time explaining technical details to notorious
crackpots as you
Tom Roberts
> Tom Roberts wrote on Fri, 12 Sep 2008 16:42:02 -0500:
>> Note that I agree that "GR does not reduce to NG"
>
This is basically a quibble about the meaning of the word "reduce". Some
(like you, and I quoting you above) use it to mean a series of theorems
leading from one to the other; others use it to mean an approximation
(which is of course not rigorous in the same sense).
Similarly, the few actual statements you have made are overly strong,
such as the claim that in a flat manifold GR has no "gravitational
force" -- while that is correct, it is irrelevant: NG is not GR in flat
spacetime, it is APPROXIMATELY equivalent to GR for very small fields in
a spacetime that can be foliated with a LOCALLY-flat spatial submanifold
that is identified with the space used in NG. (Note also that GR->NG is
rather more complicated than a simple c->infinity limit.)
You repeatedly make stronger statements about "the Newtonian
approximation" than are reasonable. For instance:
> My works shows, in a rigorous mathematical sense, that no metric theory
> of gravity can reproduce Newtonian results in the *complete* space
> dynamics (i.e. for arbitrary values of parameters).
The "*complete* space" is not important. What matters is that the
predictions of GR be experimentally indistinguishable from the
predictions of NG, IN DOMAINS WHERE BOTH HAVE BEEN APPLIED. That is, it
is required that the experiments that established NG do not refute GR.
This is basically here on earth and most observations in the solar
system. The rest of the "*complete* space" is irrelevant, especially if
you mean chaotic system (which appears to be the case).
> It is also true that in that *specific* case, Schwarzschild
> metric, GR trajectories are practically 'indistinguishable' [4] from NG
> ones for low values of (M/R).
Right! This statement of yours is saying what I have been saying, just
phrased differently. This is basically what people mean by the Newtonian
limit of GR. Not the much stronger claims you make.
> But, and this is the point where the referee *fails*, the difference
> between orbits is very important in many-body situations, which are
> characteristically chaotic.
This is not the referee's failure, it is YOURS. You have made vague
reference to claims like this, but have never substantiated them. This
is a rather surprising claim, and the referee rightly expects YOU to
substantiate it. As do I.
"Extraordinary claims require extraordinary proof."
-- James "The Amazing" Randi
For a chaotic system, I would certainly expect NG and GR to yield
different orbits, when compared over a long-enough time. That's what
"chaos" means.
> In this more general case, differences of order (M/R), no matter how
> small, are *exponentially* amplified (the unitary propagator of the
> dynamics is e^Lt) with the result that GR solution turns unstable and
> does not correspond to the NG result.
Nobody would expect two theories that are only approximately the same to
give identical results for a chaotic system -- that's what "chaos"
means! Again you make an overly-strong statement. In this case I cannot
assess whether or not it is true (because you have never substantiated
it), but it's clear to me that nobody would care about the discrepancy
you claim, for discussing the Newtonian limit of GR.
As for being unstable, in GR orbits are always unstable -- among other
things they radiate gravitational radiation. But for the solar system
this instability is so small it is not observable except in certain
cases like the precession of the perihelia of inner planets. It has also
been observed for the various known binary pulsar systems.
> GR gives the wrong multiparticle correlations for a generic many-body
> state [5].
Hmmm. You need to substantiate this claim.
> In the basis of that result, it is not strange that worldwide expertises
> on relativistic chaos as Prof. W. C. Schieve
> http://order.ph.utexas.edu/research/glimpse.html
> have *abandoned* GR for the difficult task of developing a relativistic
> many body dynamics:
> http://order.ph.utexas.edu/mtrump/manybody/
Interesting.... But for test particles orbiting a mass, probability
distributions are not needed, and trajectories are highly accurate. But
Jupiter is not really a "test particle"....
From these webpages alone, it's not clear to me that this
is really a different theory, and I suspect it is basically
a new way of codifying perturbative techniques for
multi-body chaotic systems. That, of course, does not make
it uninteresting, but it is interest at a different level.
So here's a possible test of all this: show that over long periods of
time (e.g. the past few hundred years for which good observations are
available), the predictions of this new theory for the solar system
differ from NG, and get the correct answer. It's not clear to me that
this can be done, however, as they do model lunar eclipses thousands of
years in the past....
Just as Einstein knew he had to provide some experimental tests in his
1915 paper on GR, so too you need to do so.
In short: you are too hung up on the Newtonian limit of GR. You
repeatedly make surprising, antagonistic statements about it that are
really stronger than are reasonable, and you cannot expect other people
(referees) to accept your claims without solid substantiation. For
instance, your claim "there is no Newtonian limit of GR" is wrong,
because you use those words with different meanings than everybody else.
You also seem to be claiming that both NG and GR are wrong for chaotic
systems -- that is a completely different issue (though it seems to be
related, as that appears to be the domain where you claim GR->NG fails).
IOW: stop acting like a crackpot and start acting like an adult.
Tom Roberts
Eric Gisse
[...]
> In short: you are too hung up on the Newtonian limit of GR. You
> repeatedly make surprising, antagonistic statements about it that are
> really stronger than are reasonable, and you cannot expect other people
> (referees) to accept your claims without solid substantiation. For
> instance, your claim "there is no Newtonian limit of GR" is wrong,
> because you use those words with different meanings than everybody else.
> You also seem to be claiming that both NG and GR are wrong for chaotic
> systems -- that is a completely different issue (though it seems to be
> related, as that appears to be the domain where you claim GR->NG fails).
>
> IOW: stop acting like a crackpot and start acting like an adult.
>
> Tom Roberts
I've told him most of what you just said over the last 6 months, and
it hasn't happened yet.
I'm personally still waiting for him to publish - in an open,
accessible way, his manuscript that proves all this.
Juan R.
> Juan R. González-Álvarez wrote:
>> Tom Roberts wrote on Fri, 12 Sep 2008 16:42:02 -0500:
>>> Note that I agree that "GR does not reduce to NG"
>>
>> [...] GR textbooks, papers, and encyclopedias that *wrong* statement
>> that "GR reduces to NG".
>
> This is basically a quibble about the meaning of the word "reduce". Some
> (like you, and I quoting you above) use it to mean a series of theorems
> leading from one to the other;
You insist on that PUN. Reduction has a well-known standard meaning:
(\blockquote
simplify the form of a mathematical equation of expression by
substituting one term for another
)
Or more rigorous:
(\blockquote
The process of transforming an expression according to certain reduction
rules. The most important forms are beta reduction (application of a
lambda abstraction to one or more argument expressions) and delta
reduction (application of a mathematical function to the required number
of arguments).
)
In the basis of that standard meaning, and some recent research, I
conclude that GR does *not* reduce to NG; this is contrary to usual claim.
> (Note also that GR->NG is
> rather more complicated than a simple c->infinity limit.)
Once again you are confounding readers.
I am *not* the naive theoretician who is doing that kind of claims, but
General Relativists (Ehlers, Carlip...) are doing it. For instance, a
typical General Relativist claim is:
http://relativity.livingreviews.org/open?
pubNo=lrr-2005-6&page=articlesu34.html
(\blockquote
Most textbooks on general relativity discuss the fact that Newtonian
gravitational theory is the limit of general relativity as the speed of
light tends to infinity.
)
>> My works shows, in a rigorous mathematical sense, that no metric theory
>> of gravity can reproduce Newtonian results in the *complete* space
>> dynamics (i.e. for arbitrary values of parameters).
>
> The "*complete* space" is not important. What matters is that the
> predictions of GR be experimentally indistinguishable from the
> predictions of NG, IN DOMAINS WHERE BOTH HAVE BEEN APPLIED.
*Complete* space dynamics refers to whole range of Newtonian dynamical
spectra.
> For a chaotic system, I would certainly expect NG and GR to yield
> different orbits, when compared over a long-enough time. That's what
> "chaos" means.
False. The time depends on how strong dynamical correlations are; it may
be years or just minutes...
>> GR gives the wrong multiparticle correlations for a generic many-body
>> state [5].
>
> Hmmm. You need to substantiate this claim.
Sure I will do, but in research *papers* and for people that at least can
understand the notation.
Could you write the equation of motion for a two body correlation, say in
Newtonian regime up to lambda^2 order?
Well, I know that your response is "No", then the question is why would I
waste my time explaining basic stuff to arrogant people who likes to
insult and lie?
>> In the basis of that result, it is not strange that worldwide
>> expertises on relativistic chaos as Prof. W. C. Schieve
>> http://order.ph.utexas.edu/research/glimpse.html have *abandoned* GR
>> for the difficult task of developing a relativistic many body dynamics:
>> http://order.ph.utexas.edu/mtrump/manybody/
>
> Interesting.... But for test particles orbiting a mass, probability
> distributions are not needed, and trajectories are highly accurate.
As usual you are saying nothing new. From the above reference:
(\blockquote
Current [classical relativistic mechanics] theory allows the solution for
equations of motion only in the case of particles interacting with an
external field, or with a stationary particle of infinite source mass.
)
The interesting case, is of course, when the current theory fails.
(snip)
> In short: you are too hung up on the Newtonian limit of GR. You
> repeatedly make surprising, antagonistic statements about it that are
> really stronger than are reasonable, and you cannot expect other people
> (referees) to accept your claims without solid substantiation.
It is not the case with this 'anonymous' [#] referee from top GR journal.
He first rejected the paper in the basis of "this work adds nothing new to
literature on Newtonian limits".
Which is plain false for anyone who read my paper. In my reply I invited
referee to provide a single reference containing previous work on my
research.
Of course he never provided me one and changed the report to "some parts
of the work are known [##] and others are wrong".
Again I complained to editors because referee was doing too strong claims
could be not checked.
In a last report, the referee indicated some parts he thinks are wrong.
What disappointment!
For instance, in one part of his report he states that my analysis of
cosmological problems with functions of kind F(r,t) is right and point
that same conclusions hold for functions G(r(t)).
Therefore he concludes that my criticism of GR boundaries (I show that
Ehlers works on Newtonian limits is wrong see living reviews above) also
applies to the alternative solution I am proposing.
I agree with referee that functions G would be also problematic, but my
paper deal with functions of kind H(R(t)) and those new functions are
free of the problem discussed.
Therefore, I am very happy *that* report was all the criticism that a
recognized expertise on GR can do on my novel work.
Now I am busy ending another three papers. When less busy I will
improve the draft on Newtonian limits and reply referee last report.
(sniped an ad hominem)
[#] After several complaints one of assistant editors gave me referee
identity. He is an expertise in numerical methods in GR. He became
famous because proposed a Star Trek like engine. An engine would not
work according to my rigorous analysis of GR equations of motion :-)
[##] In one section of my work I analize Carroll derivation of Newtonian
equation of motion and show that his derivation is wrong. In the weak
field limit the equation of motion is (a = 0) and not the equation
Carroll gives in his lecture notes in Arxiv.
Referee agrees with me that acceleration is exactly zero for the
metric that Carroll uses.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Juan R.
Some clarifications.
> I am *not* the naive theoretician who is doing that kind of claims, but
> General Relativists (Ehlers, Carlip...) are doing it. For instance, a
> typical General Relativist claim is:
>
> http://relativity.livingreviews.org/open?
> pubNo=lrr-2005-6&page=articlesu34.html
>
> (\blockquote
> Most textbooks on general relativity discuss the fact that Newtonian
> gravitational theory is the limit of general relativity as the speed of
> light tends to infinity.
> )
This General Relativists' claim is plain wrong.
> Therefore he concludes that my criticism of GR boundaries (I show that
> Ehlers works on Newtonian limits is wrong see living reviews above) also
> applies to the alternative solution I am proposing.
I.e. referee agrees with me that boundaries used by Ehlers, Carlip, etc.
for the Newtonian limit issue are unphysical (clearly invalidated by
observation) but referee pretends to defend GR by claiming that problem
with GR solutions also applies to other alternative solutions.
Funny, specially when the problem of unphysical boundaries is absent in
the *new* solutions I have provided.
--
http://www.canonicalscience.org/en/miscellaneouszone/guidelines.html
Simple Simon
When you say "reduced circumference r", to what reduced circumference do you
refer (is not r a radial distance)?
Thank you.
Simple Simon
Never mind. Thanks.
Dirk Van de moortel
No problem. It's terminology from Taylor and Wheeler.
Radial distance cannot be measured or even conceived, due
to the 1/(1-R/r) factor in the metric.
Integrating an angle part of the metric shows that circumference
measured along the circle is 2 pi r, so r is the circumference
reduced by a factor 2 pi.
Dirk Vdm
glird
> On Sep 10, 7:31 am, "Ken S. Tucker" <dynam...@vianet.on.ca> wrote:
> > > Daryl McCullough <stevendaryl3...@yahoo.com> wrote
>
> > > >> Writing R = G M/c^2, and using accent-notation for differentiation
> > > >> w.r.t. proper time ( i.o.w. dX / d\Tau = X' ), the metric gives you
> > > >> the Lagrangian (formally divide the metric by (d\Tau)^2)
> > > >> L = c^2 = (1-R/r) c^2 t'^2 - r'^2 / (1-R/r) [1]
"L = c^2" (That's what I was looking for.)
snip
> > If anyone has Weinberg's "Grav&Cosmo", the
> > Lagrangian is well explained beginning on
> > pg.220. An interesting equation is simply,
> > dtau/dt = 1 - L , Eq.(9.2.2),
> > Ken S. Tucker
>
< This could be fun. Let me call dtau = ds, then with
a bit of calculus find, (using Eq.(9.2.2) above),
d^2(t) = dL ds , d^2(s)= -dL dt
and see how the time paradox develops using
the Lagrangian "L", in those. >
Remember, Ken, L = c^2.
(If we let c = 1 unit length/second,
in which a unit length is the distance light travels in a vacuum
in one second, then L = c^2 = 1.)
< The "ds" can be thought of as the uniform rate
at which a clocks ticks in CS k and likewise
for dt in CS K, and the d^2(t) and d^2(s) are
the accelerations of those clock *rates* as
"defined" by the change in the Lagrangian "dL". >
You wrote above, "If anyone has Weinberg's
"Grav&Cosmo", the Lagrangian is well explained
beginning on pg.220. An interesting equation is simply,
dtau/dt = 1 - L, Eq.(9.2.2)".
As with so very many others, "calculus" is
blindfolding you here, Ken. The expression dtau/dt
isn't calculus, it's just an algebraic one in which, as
you say, ds = dtau is the rate at which clocks tick in
cs k and dt is the rate at which they beat in cs K.
Therefore, dtau/dt is the ratio of rates of the two
systems, as plotted by K.
However, if "dtau/dt = 1 - L and "L = 1", then
dtau/dt = 1 - 1 = 0, which is absurd.
In the LTE, dtau/dt = sqrt(1 - v^2/c^2) = q.
If we set v = .6c, then q = .8, wherefore
dtau = qdt = .8dt; which means, If a K clock
beats off 10 seconds, a k clock will have beat off
8 seconds. Why? Because in STR. clocks of a moving
system run q slower than those of a differently moving
system; in which v is the velocity of either cs as plotted
by the other esynched (see below) system.
< Correct me if I'm wrong: In Newton Theory "L"
is constant so that dL=0 and dt^2 =0 because
both dt and ds are sync'd constant...tick, tick. >
In STR, dt (the time of a stationary system K) IS
constant. But in the esynched moving system k, ds is
a function of dx', in which ds is the difference in settings
of two k clocks a distance dx' apart as measured by K.
< If I understand correctly, there is a subtle
variation in the Lagrange "L" in transforming
from Newton's theory to GR, where in GR, L is
not constant. Is that right? >
Not quite. Even in GR, c is a constant in a vacuum.
Thus so is L = c^2. In both NR and GR, light will
pass differently moving systems at variable relative
speeds. BUT!! In GR clocks are to be esynched (set by
Descartes-Einstein operational method) whereupon they will
MEASURE c as a constant all directions.
All of this stuff is very simple once you see that
Einstein was doing algebra rather than calculus in his
1905 paper. As indicated in a different thread,
"Why relativists don't understand Einstein's 1905 mathematics",
those who DON'T know that get lost in both SR and GR; and
Physics and reality, too.
glird
Dono
> All of this stuff is very simple once you see that
> Einstein was doing algebra rather than calculus in his
> 1905 paper.
No, Lebau
Einstein was doing BOTH algebra and calculus in his 1905 paper.
The fact that you are an old fart that didn't get past 10-th grade
doesn't mean that Einstein was only doing algebra. Besides, what is
this to you? You don't understand either, Lebau.
Dirk Van de moortel
427dfc2f-7117-4a58...@r66g2000hsg.googlegroups.com
[snip tucker crap]
> As with so very many others, "calculus" is
> blindfolding you here, Ken. The expression dtau/dt
> isn't calculus, it's just an algebraic one in which, as
> you say, ds = dtau is the rate at which clocks tick in
> cs k and dt is the rate at which they beat in cs K.
> Therefore, dtau/dt is the ratio of rates of the two
> systems, as plotted by K.
We have clock object undergoing constant proper acceleration a.
In K, its proper time tau can be tracked as a function of coordinate
time t:
tau(t) = c/a arcsinh(a t/c)
How, *without* calculus, are you going to express the quantity
dtau/dt = sqrt(1-v^2/c^2)
as a function of t and a?
Dirk Vdm
glird
> glird <gl...@aol.com> wrote [in reply to Ken
Tuckers' remarks concerning the clock paradox:
>
<< As with so very many others, "calculus" is
blindfolding you here, Ken. The expression
dtau/dt isn't calculus, it's just an algebraic
at which clocks tick in cs k and dt is the rate at
which they beat in cs K.
Therefore, dtau/dt is the ratio of rates of the
two systems, as plotted by K. >>
< We have clock object undergoing constant proper
acceleration a. >
The clock paradox arose from considerations wrt
inertial systems, independently of accelerations.
<In K, its proper time tau can be tracked as a function
of coordinate time t: tau(t) = c/a arcsinh(a t/c)>
In Einstein's STR paper, tau denoted the time of
the moving system k and t denoted the time of the
stationary system K. Since (to him) "time" is nothing
but the position of the hands of a clock, co-ordinate
time is the ONLY kind of time that exists in his and
anyone's equations. As to "proper time", to Lorentz
that denoted the time (t) of the origin clock of a given
system. The time per any other clock in the direction of
motion was the "local time" (t-vx/c^2) relative to that one.
< How, *without* calculus, are you going to express the quantity
dtau/dt = sqrt(1-v^2/c^2)
as a function of t and a? >
In his 1905 STR paper (where the expression
dtau/dt comes from), Einstein substituted the
letter a for dtau/dt and then said, "a is a
function phi(v) at present unknown. Later on
he said that phi(v) = 1. If we ignore the fact
that dtau/dt = 1 rules out the LTE, and that
your "a" isn't present in either STR or the LTE or
the clock paradox; then the Lorentz value of
dtau IS a function of t, being dtau = qdt (in
which dtau denotes one second as beat of by k
clocks, dt represents that same period as beat
off by K clocks, and q = /dt = sqrt(1-v^2/c^2).
If you think that is calculus, Dirk Vdm, I
say you are wrong.
glird
Dirk Van de moortel
[snip due to clumsy quoting]
We have a clock object undergoing constant proper acceleration a.
In K, its proper time tau can be tracked as a function of coordinate
time t:
tau(t) = c/a arcsinh(a t/c)>
How, *without* calculus, are you going to express the quantity
dtau/dt = sqrt(1-v^2/c^2)
as a function of t and a?
Dirk Vdm
Ken S. Tucker
I'm lurking, with little to offer.
I think you're interested in the Lagrangian,
I'm a student of the subject, however I did
check Weinberg's work and find it reasonable.
Set L = Kinetic(T)-Potential(V) energies as
is the conventional notation.
Weinberg's "L" is approximately GM/rc^2, then
T=L+V = 2GM/rc^2.
Then find Escape Velocity = sqrt(2GM/r).
How so you find?
Ken S. Tucker
glird
>
> How, *without* calculus, are you going to express the quantity
> dtau/dt = sqrt(1-v^2/c^2)
> as a function of t and a?
In the algebra equation, dtau/dt = sqrt(1-v^2/c^2)
(in which there is no acceleration "a"), the value
of the time tau of a given moving system k is given
by clock of that system and the time t is given by
a clock of a different system K. A period of time
beat off by one such k-clock is symbolized by dtau,
and a period beat off by a K-clock by dt. In the
equation dtau/dt = sqrt(1-v^2/c^2) = q, dtau/dt is a
symbol for dtau : dt, which denotes the ratio of
rates of clocks of the two systems. Although that
ratio is a function of the velocity v of system k as
measured by K; once we arbitrarily insert a numerical
value for v the rest is simple.
Examples:
1. Let v = .8c, so q = sqrt(1-v^2/c^2) = .6
and dtau/dt = .6.
2. Let v = .6c, so q = dtau/dt = .8.
3. Let v = 250 mph.
Applications:
1. If dt = 6 seconds then dtau = qdt = .36 seconds;
so k clocks run slow compared to those of K.
2. If dt = 6 seconds then dtau = qdt = .48 seconds;
so k clocks run slower than K clocks, but not as
slow as in example 1.
3. Now, Dirk, please tell us how, *with* calculus, to
express the quantity dtau/dt = sqrt(1-v^2/c^2) as
a function of t.
glird
Dirk Van de moortel
> On Sep 25, 9:06 am, "Dirk Van de moortel" wrote:
>>
>> How, *without* calculus, are you going to express the quantity
>> dtau/dt = sqrt(1-v^2/c^2)
>> as a function of t and a?
>
> In the algebra equation, dtau/dt = sqrt(1-v^2/c^2)
> (in which there is no acceleration "a"),
and therefore I asked you:
In K, its proper time tau can be tracked as a function of coordinate
time t:
tau(t) = c/a arcsinh(a t/c)
How, without calculus, are you going to express the quantity
dtau/dt = sqrt(1-v^2/c^2)
as a function of t and a?
Here's a hint:
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
Can you do it without calculus or can't you?
Just answer the question.
Dirk Vdm
glird
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> glird <gl...@aol.com> wrote
> > On Sep 25, 9:06 am, "Dirk Van de moortel" wrote:
>
> >> How, *without* calculus, are you going to express the quantity
> >> dtau/dt = sqrt(1-v^2/c^2)
> >> as a function of t and a?
>
> > (in which there is no acceleration "a"),
>
> and therefore I asked you:
> In K, its proper time tau can be tracked as a function of coordinate
> time t: tau(t) = c/a arcsinh(a t/c)
> How, without calculus, are you going to express the quantity
> dtau/dt = sqrt(1-v^2/c^2)
> as a function of t and a?
>
> Can you do it without calculus or can't you?
> Just answer the question.
Yes.
{I already did, in the rest of the answer you snipped out.]
glird
Dirk Van de moortel
I did not see you produce an expression for dtau/dt as a
function of t and a.
You can't do it :-)
What a dope.
Dirk Vdm
Simple Simon
>> Thanks to all, I look forward to following your advice.
>>
>> Dirk Van de moortel:
>> Thank you especially for your presentation.
>>
>> [ @L/@t' ]'
>> looks to me like
>> "The derivative with respect to tau, of the Integral part of the
>> partial derivative of L with respect to the derivative of t with
>> respect to tau" I expect to learn to interpret this.
>
> No Intergral part here :-)
> "The derivative with respect to tau, of the partial derivative of L
> with respect to the derivative of t with respect to tau"
> I could have written it as
> d/d\Tau ( @L/@t' )
> but I prefer the accent.
> Yes, this is all standard Euler-Lagrange lingo.
> It's a good idea to check up on this in the context of classical
> mechanics first.
> To get some idea: http://en.wikipedia.org/wiki/Euler-Lagrange
> L is a function of the coordinates and their derivatives, written as
> L( x^m, x'^m )
> possible in ASCII, so I use accent-notation.
>
>>
>> I seem to have lost a factor of 2:
>> "
>> 2 r' r" = c^2 R r' / r^2 (we have lost the
>> constant ) or simplifying:
>> r" = c^2 R / r^2
>> "
>
> Ah yes, sorry, another pair of typos:
> 2 r' r" = - c^2 R r' / r^2 (we have lost the
> constant :-) ) or simplifying:
> or, using R = 2 G M/c^2: (typo 2: lost factor 2)
> r'' = - G M / r^2
> or, using your notation:
> d^2r/d\Tau^2 = - G M / (r^2)
>
>
> Dirk Vdm
Again, thanks to all.
My copy of "A first course in general relativity" by Bernard F. Schutz
finally arrived. While Tom Roberts estimation of my level of sophistication
was spot on, some of the "errors of omission" in the text, while not
reducing its veracity, diminishes its educational value, by making it harder
and take longer to learn.
(e.g. on Pages 7 and 8, the description of x' axis is not symmetric with its
description of the t' axis; it describes the x' axis as "those that O'
measures to be simultaneous with the event t' = x' = 0. It doesn't describe
the t' axis as "those that O' measures to be colocal with event x'=t'=0).
(e.g. On page 8 where it refers to Fig. 1.3 it describes to events (E and R)
in the figure that don't appear there.) And I'm only up to page 9! While
these are very minor, they do add up.
In the meantime, while waiting for it to arrive, I took the opportunity to
calculate a rough estimate of the /minimum/ central mass of our local
black-hole to be:
~ M = 3.18619065157 x 1053 kg.
http://sites.google.com/site/theblackholehypothesis/
Given the mass of our observable universe, this would make Harmony a very
dirty snowball.
I think that I'll try to post this work to sci.physics.foundations.
Simple Simon
That should read:
M ~= 3.18619065157 x 10^53 kg.