Prev: Einstein "fingers" and relativity
Next: Space and Time
From: BURT on 15 Nov 2009 20:18
On Nov 15, 7:07 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
> "eric gisse" <jowr.pi.nos...(a)gmail.com> wrote in message
>
> news:hdo998$ij5$1(a)news.eternal-september.org...
>
>
>
>
>
> > Jay R. Yablon wrote:
>
> >>     In the linear approximation, the metric tensor g^uv is related to
> >> the gravitational field h^uv according to (k=sqrt(16 pi G)):
>
> > The metric tensor g{^,_)uv _is_ the gravitational field.
>
> >> g^uv = eta^uv + k h^uv  (1)
>
> >> Further, the "graviton" field psi^uv is related to h^uv according to
> >> (what is the best thing to call psi^uv, in contrast to h^uv?):
>
> >> psi^uv = h^uv - .5 g^uv h   (2)
>
> > It contains the exact same information as the metric, just expressed
> > in an
> > irrelevant but slightly different way which adds nothing to the
> > discourse.
>
> Yes, they have the same information, but in a different form.  Relevance
> depends on what you are trying to do and cannot be prejudged.
>
>
>
> >>     I would like to know what (1) and (2) become, exactly, when the
> >> gravitational fields become very strong.
>
> > They become irrelevant. The linearized limit presumes weak
> > gravitation. The
> > full answer lies back with the full nonlinear field equations.
>
> Obviously one uses the non-linear field equations.  But that is not my
> question.  The entity psi^uv is taken in quantum theory to represent the
> graviton.  Yes, it has the same information as g^uv but is a function of
> g^uv.  I am simply asking what that actual function becomes.  If you
> know how to write it with R^uv, T^uv, etc., that is fine.
>
> What this is relevant to, is that I am trying to determine what would be
> the field of integration "field" in a path integral $Dfield, for
> gravitation.  If you know that for sure and can back it up, we can skip
> all this other stuff.  There was no reason for you to start pissing.
>
> {snipped pissing}
>
> Jay.- Hide quoted text -
>
> - Show quoted text -

Gravity is round.

Mitch Raemsch
From: George Hammond on 16 Nov 2009 09:22
On Sat, 14 Nov 2009 20:50:03 -0500, "Jay R. Yablon"
<jyablon(a)nycap.rr.com> wrote:

> In the linear approximation, the metric tensor g^uv is related to
>the gravitational field h^uv according to (k=sqrt(16 pi G)):
>
>g^uv = eta^uv + k h^uv (1)
>
>Further, the "graviton" field psi^uv is related to h^uv according to
>(what is the best thing to call psi^uv, in contrast to h^uv?):
>
>psi^uv = h^uv - .5 g^uv h (2)
>
> I would like to know what (1) and (2) become, exactly, when the
>gravitational fields become very strong. I believe what happens is the
>the sqrt(-g) factor kicks in, so that (1) now becomes:
>
>sqrt(-g) g^uv = eta^uv + k sqrt(-g) h^uv (3)
>
>and that the relationship (2) stays intact. Is this so? If not, what
>are the correct relationships for gravitational field of any strength?
>
> If the above is so, then combining (2) and (3), we obtain:
>
>k sqrt(-g) psi^uv = (2-sqrt(-g)) g^uv - eta^uv, (4)
>
>which subtracts off the constant flat background eta^uv. Thus, if we
>take a variation (delta) of each side, which removes out the Minkowski
>eta^uv=constant background, (4) leads to:
>
>k delta(sqrt(-g) psi^uv) = 2 delta (g^uv) - delta (sqrt(-g) g^uv) (5)
>
>which in the linear sqrt(-g)=1 approximation in rectilinear coordinates
>leads to:
>
>k delta(psi^uv) = delta (g^uv) (6)
>
>and so up to the constant k, psi^uv and g^uv vary together. But, in the
>non-linear case, (5) seems to suggest that sqrt(-g) psi^uv has two terms
>contributing to the variation, one in relation to sqrt(-g) g^uv as in
>the linear theory, and the other in relation to delta (g^uv).
>
> Again, is this correct, and if not, what are the correct, exact
>relationships in a field of unlimited strength?
>
>Thanks,
>
>Jay
>__
>
[Hammond]
If what you're saying is true, can't you demonstrate it by
rewriting (1) as:

g^uv = eta^uv + k(h^uv + higher order terms in h^uv)

where the (unknown) series of higher order terms in h^uv is
defined so as to make g^uv de facto "exact", and then do the
same calculation you did above and show that all the h^uv
terms still subtract out regardless of what they are and
that you arrive at the same result as (5). Is that
possible?
========================================
GEORGE HAMMOND'S PROOF OF GOD WEBSITE
Primary site
http://webspace.webring.com/people/eg/george_hammond
Mirror site
http://proof-of-god.freewebsitehosting.com
HAMMOND FOLK SONG by Casey Bennetto
http://interrobang.jwgh.org/songs/hammond.mp3
=======================================
From: eric gisse on 16 Nov 2009 10:41
Jay R. Yablon wrote:
[...]

>>> I would like to know what (1) and (2) become, exactly, when the
>>> gravitational fields become very strong.
>>
>> They become irrelevant. The linearized limit presumes weak
>> gravitation. The
>> full answer lies back with the full nonlinear field equations.
>
> Obviously one uses the non-linear field equations.

It is so obvious, in fact, that I don't understand why the question was
asked.

> But that is not my
> question. The entity psi^uv is taken in quantum theory to represent the
> graviton. Yes, it has the same information as g^uv but is a function of
> g^uv. I am simply asking what that actual function becomes. If you
> know how to write it with R^uv, T^uv, etc., that is fine.

How is that question supposed to be answerable?

The entity doesn't 'become' anything because it doesn't exist any more than
the metric in the weak field limit 'becomes' something in the full nonlinear
theory.

You can obviously subtract off an arbitrary tensor off the metric tensor,
but is it meaningful?

>
> What this is relevant to, is that I am trying to determine what would be
> the field of integration "field" in a path integral $Dfield, for
> gravitation. If you know that for sure and can back it up, we can skip
> all this other stuff. There was no reason for you to start pissing.

Why is there the implied expectation that quantities derived from an
approximation to general relativity retain any of their relevance in the un-
approximated theory?

>
> {snipped pissing}
>
> Jay.

From: Jay R. Yablon on 16 Nov 2009 10:41

"George Hammond" <Nowhere1(a)notspam.com> wrote in message
news:tmm2g59ftit213efbgsh9cnkpfaata5dqk(a)4ax.com...
> On Sat, 14 Nov 2009 20:50:03 -0500, "Jay R. Yablon"
> <jyablon(a)nycap.rr.com> wrote:
>
>> In the linear approximation, the metric tensor g^uv is related to
>>the gravitational field h^uv according to (k=sqrt(16 pi G)):
>>
>>g^uv = eta^uv + k h^uv (1)
>>
>>Further, the "graviton" field psi^uv is related to h^uv according to
>>(what is the best thing to call psi^uv, in contrast to h^uv?):
>>
>>psi^uv = h^uv - .5 g^uv h (2)
>>
>> I would like to know what (1) and (2) become, exactly, when the
>>gravitational fields become very strong. I believe what happens is
>>the
>>the sqrt(-g) factor kicks in, so that (1) now becomes:
>>
>>sqrt(-g) g^uv = eta^uv + k sqrt(-g) h^uv (3)
>>
>>and that the relationship (2) stays intact. Is this so? If not, what
>>are the correct relationships for gravitational field of any strength?
>>
>> If the above is so, then combining (2) and (3), we obtain:
>>
>>k sqrt(-g) psi^uv = (2-sqrt(-g)) g^uv - eta^uv, (4)
>>
>>which subtracts off the constant flat background eta^uv. Thus, if we
>>take a variation (delta) of each side, which removes out the Minkowski
>>eta^uv=constant background, (4) leads to:
>>
>>k delta(sqrt(-g) psi^uv) = 2 delta (g^uv) - delta (sqrt(-g) g^uv)
>>(5)
>>
>>which in the linear sqrt(-g)=1 approximation in rectilinear
>>coordinates
>>leads to:
>>
>>k delta(psi^uv) = delta (g^uv) (6)
>>
>>and so up to the constant k, psi^uv and g^uv vary together. But, in
>>the
>>non-linear case, (5) seems to suggest that sqrt(-g) psi^uv has two
>>terms
>>contributing to the variation, one in relation to sqrt(-g) g^uv as in
>>the linear theory, and the other in relation to delta (g^uv).
>>
>> Again, is this correct, and if not, what are the correct, exact
>>relationships in a field of unlimited strength?
>>
>>Thanks,
>>
>>Jay
>>__
>>
> [Hammond]
> If what you're saying is true, can't you demonstrate it by
> rewriting (1) as:
>
> g^uv = eta^uv + k(h^uv + higher order terms in h^uv)
>
> where the (unknown) series of higher order terms in h^uv is
> defined so as to make g^uv de facto "exact", and then do the
> same calculation you did above and show that all the h^uv
> terms still subtract out regardless of what they are and
> that you arrive at the same result as (5). Is that
> possible?
.. . .
George,

Aside from a sqrt(-g) factor which you omit, you are exactly correct.
Your reply helped me makes sense of exactly how to use
http://en.wikipedia.org/wiki/Gravitational_wave#Mathematics.

I am reminded that you are actually quite a good and insightful
physicist, which as I have remarked before, many readers probably do not
see because they see all your "God stuff" and do not look any further.

Because the metric tensor g_uv occupies -- shall we say -- a "prominent
place" in your thinking, stay tuned to my upcoming posts. I am in the
midst of using the path integral to quantize gravitation and the field
g^uv (h^uv). You would call it "quantum God," but I will not go there.
;-)

Jay.



From: George Hammond on 17 Nov 2009 03:36
On Mon, 16 Nov 2009 10:41:28 -0500, "Jay R. Yablon"
<jyablon(a)nycap.rr.com> wrote:

>
>"George Hammond" <Nowhere1(a)notspam.com> wrote in message
>news:tmm2g59ftit213efbgsh9cnkpfaata5dqk(a)4ax.com...
>> On Sat, 14 Nov 2009 20:50:03 -0500, "Jay R. Yablon"
>> <jyablon(a)nycap.rr.com> wrote:
>>
>>> In the linear approximation, the metric tensor g^uv is related to
>>>the gravitational field h^uv according to (k=sqrt(16 pi G)):
>>>
>>>g^uv = eta^uv + k h^uv (1)
>>>
>>>Further, the "graviton" field psi^uv is related to h^uv according to
>>>(what is the best thing to call psi^uv, in contrast to h^uv?):
>>>
>>>psi^uv = h^uv - .5 g^uv h (2)
>>>
>>> I would like to know what (1) and (2) become, exactly, when the
>>>gravitational fields become very strong. I believe what happens is
>>>the
>>>the sqrt(-g) factor kicks in, so that (1) now becomes:
>>>
>>>sqrt(-g) g^uv = eta^uv + k sqrt(-g) h^uv (3)
>>>
>>>and that the relationship (2) stays intact. Is this so? If not, what
>>>are the correct relationships for gravitational field of any strength?
>>>
>>> If the above is so, then combining (2) and (3), we obtain:
>>>
>>>k sqrt(-g) psi^uv = (2-sqrt(-g)) g^uv - eta^uv, (4)
>>>
>>>which subtracts off the constant flat background eta^uv. Thus, if we
>>>take a variation (delta) of each side, which removes out the Minkowski
>>>eta^uv=constant background, (4) leads to:
>>>
>>>k delta(sqrt(-g) psi^uv) = 2 delta (g^uv) - delta (sqrt(-g) g^uv)
>>>(5)
>>>
>>>which in the linear sqrt(-g)=1 approximation in rectilinear
>>>coordinates
>>>leads to:
>>>
>>>k delta(psi^uv) = delta (g^uv) (6)
>>>
>>>and so up to the constant k, psi^uv and g^uv vary together. But, in
>>>the
>>>non-linear case, (5) seems to suggest that sqrt(-g) psi^uv has two
>>>terms
>>>contributing to the variation, one in relation to sqrt(-g) g^uv as in
>>>the linear theory, and the other in relation to delta (g^uv).
>>>
>>> Again, is this correct, and if not, what are the correct, exact
>>>relationships in a field of unlimited strength?
>>>
>>>Thanks,
>>>
>>>Jay
>>>__
>>>
>> [Hammond]
>> If what you're saying is true, can't you demonstrate it by
>> rewriting (1) as:
>>
>> g^uv = eta^uv + k(h^uv + higher order terms in h^uv)
>>
>> where the (unknown) series of higher order terms in h^uv is
>> defined so as to make g^uv de facto "exact", and then do the
>> same calculation you did above and show that all the h^uv
>> terms still subtract out regardless of what they are and
>> that you arrive at the same result as (5). Is that
>> possible?
>. . .
>George,
>
>Aside from a sqrt(-g) factor which you omit, you are exactly correct.
>Your reply helped me makes sense of exactly how to use
>http://en.wikipedia.org/wiki/Gravitational_wave#Mathematics.
>
>
[Hammond]
[Hammond]
I was not aware, and apparently Gisse isn't either, that
the EFE can be transformed (exactly) into a wave equation in
h^uv simply by defining h^uv as:

h^uv = eta^uv - sqrt(-g) g^uv

and that this well known transformation is used to describe
gravitational radiation, particularly binary star radiation,
as the Wiki page you cited above informs us.

>I am reminded that you are actually quite a good and insightful
>physicist, which as I have remarked before, many readers probably do not
>see because they see all your "God stuff" and do not look any further.
>
>
[Hammond]
Well, you can relax about the "god stuff" J.Y., in that the
"god problem" has actually been scientifically solved. It
took 25 years to find it and I never even knew where I was
going until the last couple of years. Turns out that that
the God of the Bible is completely explained by classical GR
without any appeal quantum phenomena.
>
>
>Because the metric tensor g_uv occupies -- shall we say -- a "prominent
>place" in your thinking, stay tuned to my upcoming posts. I am in the
>midst of using the path integral to quantize gravitation and the field
>g^uv (h^uv). You would call it "quantum God," but I will not go there.
>;-)
>Jay
>
[Hammond]
If classical GR explains classical God, then certainly
"quantum gravity" is going to advance our knowlege of the
God phenomenon even further and could even yield some
dramatically surprising results.
But in the meantime it will take the Church and Academia
years just to come to grips with the discovery of a
classical proof of God, never mind worrying about
quantum-god phenomena!

But in the meantime I have a question for you. Prof.
Frank Tipler (Tulane) is a world class expert in GR,
personal friend of Wolfgang Rindler, Roger Penrose, etc. In
one of his recent best selling books he says:

"Richard Feynman discovered a consistent (renormalizeable is
the technical term) theory of quantum gravity 40 years ago,
and this theory is essentially unique."

Yet such authorities as Roger Penrose will be the first to
tell you that "not only have we not discovered the theory of
quantum gravity, we don't even have a reasonable candidate
for such a theory".

What the heck is Frank Tipler talking about? I was under
the impression that Richard Feynman's 1962 solution to the
problem of quantum gravity wasn't renormalizeable?

Everyone knows that Tipler is a genius and a stellar
mathematician....... why does he keep saying that Feynman
solved quantum gravity 40 years ago when everybody else
disagrees?
========================================
GEORGE HAMMOND'S PROOF OF GOD WEBSITE
Primary site
http://webspace.webring.com/people/eg/george_hammond
Mirror site
http://proof-of-god.freewebsitehosting.com
HAMMOND FOLK SONG by Casey Bennetto
http://interrobang.jwgh.org/songs/hammond.mp3
=======================================
First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4
Prev: Einstein "fingers" and relativity
Next: Space and Time