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 $Dg_uv?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
carlip...@physics.ucdavis.edu
> If one were to use the Einstein-Hilbert action in the usual
> expression for a path integral, where the metric tensor
> g_uv (or at least its gravitational field components h_uv)
> is itself the field of interest, in order to see what quantum
> gravitational looks like, the path integral would be specified
> by (k=the usual kappa):
> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
No. Your gravitational part is right, but your matter part is not;
you need to put in the actual matter Lagrangian. (If you vary
the metric in the action you've written down, you won't get
the Einstein field equations.)
> I have written in (1), that the integration is over "$Dg."
> My question is this:
> Should this specifically be:
> $Dg_uv? (2)
> or, should this be:
> $D(sqrt(-g)g_uv)? (3)
This "measure problem" is fairly famous. It would be a lot easier to
answer if we actually had a quantum theory of gravity -- that is, if
we knew that the path integral was well-defined, or at least well-
enough-defined to let us do sensible calculations. Since we don't,
any reply to your question is little more than a guess.
You will find one attractive proposal -- and references to others --
in Bern, Blau, and Mottola, Phys. Rev. D43 (1991) 1212. You might
also look at the new preprint by Han and Thiemann, arXiv:0911.3428,
for a general (but nonrigorous) discussion of how to obtain the measure.
(If you want rigor, look at Cartier and DeWitt-Morette's book,
_Functional Integration_.)
Steve Carlip
Jay R. Yablon
news:he44uv$h5l$1...@skeeter.ucdavis.edu...
> Jay R. Yablon <jya...@nycap.rr.com> wrote:
>> If one were to use the Einstein-Hilbert action in the usual
>> expression for a path integral, where the metric tensor
>> g_uv (or at least its gravitational field components h_uv)
>> is itself the field of interest, in order to see what quantum
>> gravitational looks like, the path integral would be specified
>> by (k=the usual kappa):
>
>> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
>
> No. Your gravitational part is right, but your matter part is not;
> you need to put in the actual matter Lagrangian. (If you vary
> the metric in the action you've written down, you won't get
> the Einstein field equations.)
Dear Dr. Carlip,
Thank you for your comments. I respectfully disagree. Equation (1)
above *is* consistent with the Einstein field equations. But, as I now
have come to realize, this is apparently not known.
In the two-section, <10 page discussion linked below, I have in the
first section (just over 3 pages) laid out a proof that the
Einstein-Hilbert action can indeed be expressed as written above, and is
consistent with the Einstein equation.
http://jayryablon.files.wordpress.com/2009/11/einstein-hilbert-action-4.pdf
I would ask you to please take a close look at this proof and let me
know whether you come to agree with me.
I realized from reading your comments, as well as from "X-Phy" at
http://groups.google.com/group/sci.physics.research/browse_frm/thread/bf59203692ad7033#,
and also from Ilja Schmelzer at
http://groups.google.com/group/sci.physics.foundations/browse_frm/thread/a396e0d58f0498c1#,
that (1) above is apparently *not* a known relationship. I had proven
(1) for myself as an exercise, and had been using this action in my own
pursuits, unaware of the fact that this is apparently not known. I now
see that it apparently is unknown, and suspect that this may be why it
has been so difficult to lay a well-defined foundation for doing path
integrals for gravitational theory.
>
>> I have written in (1), that the integration is over "$Dg."
>
>> My question is this:
>> Should this specifically be:
>
>> $Dg_uv? (2)
>
>> or, should this be:
>
>> $D(sqrt(-g)g_uv)? (3)
>
> This "measure problem" is fairly famous. It would be a lot easier to
> answer if we actually had a quantum theory of gravity -- that is, if
> we knew that the path integral was well-defined, or at least well-
> enough-defined to let us do sensible calculations. Since we don't,
> any reply to your question is little more than a guess.
I thank you for the papers below, and I did buy and download Bern, Blau,
and Mottola and found it helpful in clarifying my own thinking. I also
obtained Han and Thiemann and found a number of related arxiv papers.
Once the relationship Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv
((1/k)R^uv-T^uv) stated above is shown to be correct, the possibilities
for selecting a suitable "measure" become much more clearly defined.
Section 2 in the above
http://jayryablon.files.wordpress.com/2009/11/einstein-hilbert-action-4.pdf
lays this out, and shows a path integral that is "well-defined, or at
least well enough-defined to let us do sensible calculations." If the
proof in section 1 is sustained by your review, I would ask you to then
please take a look at section 2.
Thanks, and best regards,
Jay R. Yablon
Jay R. Yablon
news:7moue3F...@mid.individual.net...
> <carlip...@physics.ucdavis.edu> wrote in message
> news:he44uv$h5l$1...@skeeter.ucdavis.edu...
>> Jay R. Yablon <jya...@nycap.rr.com> wrote:
>>> If one were to use the Einstein-Hilbert action in the usual
>>> expression for a path integral, where the metric tensor
>>> g_uv (or at least its gravitational field components h_uv)
>>> is itself the field of interest, in order to see what quantum
>>> gravitational looks like, the path integral would be specified
>>> by (k=the usual kappa):
>>
>>> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
>>
>> No. Your gravitational part is right, but your matter part is not;
>> you need to put in the actual matter Lagrangian. (If you vary
>> the metric in the action you've written down, you won't get
>> the Einstein field equations.)
>
> Dear Dr. Carlip,
>
> Thank you for your comments. I respectfully disagree. Equation (1)
> above *is* consistent with the Einstein field equations. But, as I
> now
> have come to realize, this is apparently not known.
I stand corrected. Equation (1) above is wrong, because as Dr.
Carlip pointed out to me privately, the traceless EM field would
contribute nothing in (1), which is physically incorrect. The specific
mathematical error in the derivation I posted yesterday, was that even
though the trace of the Einstein equation kT=R, I was applying this in a
variational situation, as k delta T = delta R, where it is not true.
Keeping in mind Edison's saying that every success he had was
accompanied by a thousand failures, and seeing every error as an
opportunity to learn, I now understand that aside from the "measurement
problem," the difficulty in calculating path integrals with the Einstein
Hilbert action resides in the fact that there is no direct occurrence of
the source energy tensor T^uv or its trace T in the E-H action, and so
one does not have the ability to use a variation delta/delta T to remove
terms from under the integral, as one can do with delta/delta J in gauge
theory.
So, something else is needed to evaluate the *corrected*
Einstein-Hilbert path integral with the matter Lagrangian:
Z=${-oo to +oo}Dg exp[i $d^4x sqrt(-g)((1/2k)R+L_m)] (2)
I think I may have finally figured out how to do this correctly. In
a new exercise linked at:
I have endeavored to show, mathematically, how to evaluate the path
integral (2) for the Einstein-Hilbert action. In section 1, we briefly
review the Einstein-Hilbert action and insert this into the standard
Feynman path integral. That is, in section 1, we simply obtain (2),
while laying out the standard, known background material which supports
this action and relates it to the Einstein equation. Nothing in section
1 is new.
Then, in section 2, we show how to *reparameterize* the matter
Lagrangian in such a way as to allow the path integral (2) to be
mathematically evaluated over the definite field-density range from
negative to positive infinity, using Fourier analysis.
The end result, deduced in (2.25) and generalized for various choices of
"measure" in (2.33), is the *mathematical* deduction that the evaluation
of (2) above over the definite range -oo to +oo is as follows:
Z=${-oo to +oo} D(sqrt(-g)g_ab) exp [i $d^4x sqrt(-g) ((1/2k)R + L_m)]
= delta^(4xoo)[.25 g_ab ((1/2k)R + L_m) ] (3)
where delta^(4xoo) is a 4 x infinite dimensional Dirac delta (Fourier
impulse) function.
The E-H Lagrangian density:
L = (1/2k)R + L_m (4)
moves through the path integration intact, and ends up inside an impulse
function in the space conjugate to spacetime. In this formulation,
traceless EM field (L=.25 F^uv F_uv) clearly would contribute.
I want to be very clear on one point: I am saying nothing here or in
this paper about the physics that is associated with (3). I am simply
addressing the *mathematical* problem of evaluating the definite
integral (2), and getting a real answer, which, according to what is
deduced here, the mathematics tells us is (3). The only place where
anything other than pure mathematical calculation comes into play, is in
selecting the "measure" for the integration field. Rather than "choose"
one measure over another, I have used what seem to be the four most
reasonable possibilities for the measure, and, as it turns out, they all
lead to the same mathematical outcome.
I will keep my fingers crossed that perhaps I finally have hunted
this down the right way?
Thanks.
Jay
Igor Khavkine
> <carlip-nos...@physics.ucdavis.edu> wrote in message
> news:he44uv$h5l$1...@skeeter.ucdavis.edu...
> > Jay R. Yablon <jyab...@nycap.rr.com> wrote:
> >> If one were to use the Einstein-Hilbert action in the usual
> >> expression for a path integral, where the metric tensor
> >> g_uv (or at least its gravitational field components h_uv)
> >> is itself the field of interest, in order to see what quantum
> >> gravitational looks like, the path integral would be specified
> >> by (k=the usual kappa):
>
> >> Z(g)=$Dg exp.5i $d^4x sqrt(-g)g_uv ((1/k)R^uv-T^uv) (1)
>
> > No. Your gravitational part is right, but your matter part is not;
> > you need to put in the actual matter Lagrangian. (If you vary
> > the metric in the action you've written down, you won't get
> > the Einstein field equations.)
> Thank you for your comments. I respectfully disagree. Equation (1)
> above *is* consistent with the Einstein field equations. But, as I now
> have come to realize, this is apparently not known.
That's false. What is known is that, as Steve Carlip pointed out, does
not reproduce the desired Einstein equations.
> In the two-section, <10 page discussion linked below, I have in the
> first section (just over 3 pages) laid out a proof that the
> Einstein-Hilbert action can indeed be expressed as written above, and is
> consistent with the Einstein equation.
>
> http://jayryablon.files.wordpress.com/2009/11/einstein-hilbert-action-4.pdf
>
> I would ask you to please take a close look at this proof and let me
> know whether you come to agree with me.
You make an elementary mistake between equations (1.13) and (1.15).
Your substitution of the Ricci scalar for the stress-energy trace is
illegitimate. That is only allowed on shell (for particular solutions
of Einstein equations), while you intend to make the substitution in
the action, where Einstein equations have not yet been applied, g and
T are unrelated and unrestricted. The fact that your conclusion is
incorrect is simply checked by looking at your own equation (1.9). If
L_M = trace(T), where T is the same as on the left hand side, the
equality clearly does not hold.
Igor
Igor Khavkine
> Z=${-oo to +oo} D(sqrt(-g)g_ab) exp [i $d^4x sqrt(-g) ((1/2k)R + L_m)]
> = delta^(4xoo)[.25 g_ab ((1/2k)R + L_m) ] (3)
>
> where delta^(4xoo) is a 4 x infinite dimensional Dirac delta (Fourier
> impulse) function.
This result cannot be correct, for elementary reasons. One is the
obvious fact that your right hand side depends on g_ab, which you have
supposedly integrated out. The second reason, is that you can only get
a (generalized) delta function from an integration such as above if
the argument of the exponent is linear in the quantity you are
integrating with respect to. In your case, that argument is clearly
not linear.
> The E-H Lagrangian density:
>
> L = (1/2k)R + L_m (4)
>
> moves through the path integration intact, and ends up inside an impulse
> function in the space conjugate to spacetime.
Sorry, but the last statement does not even make sense.
Igor
P.S.: Also, please do not spam the newsgroup with multiple, nearly
identical posts. If you wish to bring attention to something, start a
new thread.
Jay R. Yablon
news:663c34c1-31ba-4720...@a21g2000yqc.googlegroups.com...
> On Nov 22, 8:51 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>
>> Z=${-oo to +oo} D(sqrt(-g)g_ab) exp [i $d^4x sqrt(-g) ((1/2k)R +
>> L_m)]
>> = delta^(4xoo)[.25 g_ab ((1/2k)R + L_m) ] (3)
>>
>> where delta^(4xoo) is a 4 x infinite dimensional Dirac delta (Fourier
>> impulse) function.
>
> This result cannot be correct, for elementary reasons. One is the
> obvious fact that your right hand side depends on g_ab, which you have
> supposedly integrated out. The second reason, is that you can only get
> a (generalized) delta function from an integration such as above if
> the argument of the exponent is linear in the quantity you are
> integrating with respect to. In your case, that argument is clearly
> not linear.
The right hand side contains .25 g_ab because on the left, we can
introduce a 1=.25 g_ab g^ab and then match up the argument of the
exponent linearly with respect to the quantity against which the
intergation occurs, so the actual math of getting an impulse was not
wrong here. Equation (2.17) from what I linked at
http://jayryablon.files.wordpress.com/2009/11/fourier-path-integration-of-the-einstein-hilbert-action.pdf
shows how I did this.
*Nonetheless, this is all moot.* I thought about the g_ab being in the
RHS which is essentially the same issue you raise, and concluded by
myself that *it is wrong that there be a g_ab in the RHS at all*, and so
agree with you entirely on this point. I updated this two days ago to
fix exactly what you point out, in a link at:
And then, last night, I prepared a further, much simplified update at:
*I agree with you that g_ab cannot be in the measure, and certainly
cannot end up in the RHS after evaluating the definite integral.* The
measure which does work to overcome this problem is just sgrt(-g) by
itself. As pointed out by Bern, Blau, and Mottola, General covariance
of the path integral for quantum gravity, Phys. Rev. D43 (1991) 1212
(thanks to Dr. Carlip for the reference earlier in the thread), earlier
investigations by by Fradkin and Vilkovisky "concluded that the measure
must contain apparently noncovariant factors" including sqrt(-g). The
basic idea above of having a non-covariant measure be the variable of
integration so it disappears following the definite integration, I
respectfully submit, remains useful and is not flawed. But the
execution you discuss above was flawed, and that is corrected by using
only sgrt(-g) in the later two links. Because I discuss this at length
in other posts not yet up, and in adherence to your wishes, I will
refrain from a duplicate discussion and will rely on those other posts
for more details.
>> moves through the path integration intact, and ends up inside an
>> impulse
>> function in the space conjugate to spacetime.
>
> Sorry, but the last statement does not even make sense.
Whenever you do a Fourier transform or inverse transform, you are moving
between one "space" and another "space." e.g., ordinary space and
momentum space. Perhaps I was clumsy there (and I did mis-speak because
I should have referred to the "field" space not "spacetime"), but that
is all I was saying.
Jay