What would be the field element for Path Integation of the Einstein - Hilbert Action?
Jay R. Yablon
for a path integral, where the metric tensor g_uv (or at least its
gravitational field components h_uv) is itself the field of interest, in
order to see what quantum gravitational looks like, the path integral
would be specified by (k=the usual kappa):
Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
I have written in (1), that the integration is over "$Dg."
My question is this:
Should this specifically be:
$Dg_uv? (2)
or, should this be:
$D(sqrt(-g)g_uv)? (3)
I say this a) noting that sqrt(-g) must be a part of every integral over
spacetime, and b) having done some downstream calculation from (1) it
seems to fit everything together much better if the answer is really (3)
rather than (2). Thus, for scalar fields, we take $Dpsi, and for vector
fields we take $DA_u. But, for spin-2 fields, where the fields also are
part and parcel with the geometry and directly impact the volume
elements, do we end up taking $D(sqrt(-g)g_uv) rather than just
analogizing up from spin-1 to $Dg_uv?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Ilja
> If one were to use the Einstein-Hilbert action in the usual expression
> for a path integral, where the metric tensor g_uv (or at least its
> gravitational field components h_uv) is itself the field of interest, in
> order to see what quantum gravitational looks like, the path integral
> would be specified by (k=the usual kappa):
>
> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) � (1)
A strange way to write this. The integral in the path integral
formalism is over the Lagrangian, which is
(R-2/\)sqrt{-g} + L_matter. Sometimes this may be fine, when
sqrt{-g} g_uvT^uv = L, but in general?
> I have written in (1), that the integration is over "$Dg."
>
> My question is this:
>
> Should this specifically be:
>
> $Dg_uv? � (2)
>
> or, should this be:
>
> $D(sqrt(-g)g_uv)? � �(3)
I can suggest some reasons to prefer the variables
f^uv = sqrt(-g)g^uv (note upper indices)
in comparison with the original g_uv variables: This holds in
particular if you think about "field components" h_uv: This suggests
a theory with independent background, and on this background
the harmonic gauge d_u f^uv = 0 is extremely important.
But I don't know a good reason to consider sqrt(-g)g_uv,
whatever the context.
Jay R. Yablon
"Ilja" <ilja.sc...@googlemail.com> wrote in message
news:8772f8b8-f0c0-4f41...@g22g2000prf.googlegroups.com...
> On 10 Nov., 10:39, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> If one were to use the Einstein-Hilbert action in the usual
>> expression
>> for a path integral, where the metric tensor g_uv (or at least its
>> gravitational field components h_uv) is itself the field of interest,
>> in
>> order to see what quantum gravitational looks like, the path integral
>> would be specified by (k=the usual kappa):
>>
>> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
>
> A strange way to write this. The integral in the path integral
> formalism is over the Lagrangian, which is
> (R-2/\)sqrt{-g} + L_matter. Sometimes this may be fine, when
> sqrt{-g} g_uvT^uv = L, but in general?
Neglecting the cosmological constant /\, we have:
L = (1/2 kappa) R + L_matter
In what circumstance would L_matter *not* be a constant times the trace
matter density T?
>> I have written in (1), that the integration is over "$Dg."
>>
>> My question is this:
>>
>> Should this specifically be:
>>
>> $Dg_uv? (2)
>>
>> or, should this be:
>>
>> $D(sqrt(-g)g_uv)? (3)
>
> I can suggest some reasons to prefer the variables
>
> f^uv = sqrt(-g)g^uv (note upper indices)
>
> in comparison with the original g_uv variables: This holds in
> particular if you think about "field components" h_uv: This suggests
> a theory with independent background, and on this background
> the harmonic gauge d_u f^uv = 0 is extremely important.
>
> But I don't know a good reason to consider sqrt(-g)g_uv,
> whatever the context.
>
I have not taken a close look at whether the indexes should be upper or
lower yet, but will do so.
My real point was whether the integration variable should be g_uv (or
g^uv) alone, or whether it is preferred that one of these be multiplied
by sqrt(-g) which was my thinking.
On that question, I take it that you and I agree, or that you at least
do not contradict the thinking, that sqrt(-g) should be included in the
integration variable?
Thanks,
Jay
Jay R. Yablon
"Ilja" <ilja.sc...@googlemail.com> wrote in message
news:8772f8b8-f0c0-4f41...@g22g2000prf.googlegroups.com...
> I can suggest some reasons to prefer the variables
>
> f^uv = sqrt(-g)g^uv (note upper indices)
>
> in comparison with the original g_uv variables: This holds in
> particular if you think about "field components" h_uv: This suggests
> a theory with independent background, and on this background
> the harmonic gauge d_u f^uv = 0 is extremely important.
>
> But I don't know a good reason to consider sqrt(-g)g_uv,
> whatever the context.
>
I have been looking at:
I can follow the whole calculation, except I am banging my head on the
wall to try to figure out, under "alternative form," where they get the
final term on the first line:
-g^uv Gamma^s_sp sqrt(-g)
Are they using g^uv_;p = 0 in some unstated way? I only see that there
should be two connection terms, because there are two indexes, but
apparently the sqrt(-g) inside the covariant derivative causes a third
term, and perhaps this has to do with sqrt(-g) not transforming as a
scalar? If so, how, exactly?
Help!
Thanks,
Jay
Jay R. Yablon
"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message
news:7m0t24F...@mid.individual.net...
It is because:
d_;v sqrt(-g) = 0
but:
d_v sqrt(-g) = sqrt(-g) Gamma ^s_vs
Jay.
Jay R. Yablon
. . .
> I got it!
>
> It is because:
>
> d_;v sqrt(-g) = 0
>
> but:
>
> d_v sqrt(-g) = sqrt(-g) Gamma ^s_vs
>
> Jay.
this. Jay.
Ken S. Tucker
> "Jay R. Yablon" <jyab...@nycap.rr.com> wrote in messagenews:7m115iF...@mid.individual.net...
> . . .> I got it!
>
> > It is because:
>
> > d_;v sqrt(-g) = 0
>
> > but:
>
> > d_v sqrt(-g) = sqrt(-g) Gamma ^s_vs
>
> > Jay.
>
> ... and I just made a change to the Wiki page so others will not miss
> this. Jay.
Hi Jay,
Speigel's "Vector Analysis" has that as a 'supplementary problem'
chapter 8, #45(c), pg 192, for additional reference, (Fred has that
ref too).
In terms of the 2nd kind of Christoffel, he writes,
{j,jm} = &/&x^m ln sqrt(g) = (1/2g) &g/&x^m.
Same ref, pg.174, he advises writing sqrt(g) as sqrt(|g|) to include
g<0 or g>0, depending on the CS you choose.
Regards
Ken S. Tucker