Exercise attempting to evaluate the path integral of the Einstein-Hilbert action over the range from -00 to +00
Jay R. Yablon
exercise where I have endeavored to show, mathematically, how to
evaluate the path integral:
Z=${-oo to +oo}Dg exp[i $d^4x sqrt(-g)((1/2k)R+L_m)] (1)
for the Einstein-Hilbert action. This is at the link below:
In section 1, we briefly review the Einstein-Hilbert action and insert
this into the standard Feynman path integral. That is, in section 1, we
simply obtain (1), while laying out the standard, known background
material which supports this action and relates it to the Einstein
equation. Nothing in section 1 is new.
Then, in section 2, we show how to *reparameterize* the matter
Lagrangian in such a way as to allow the path integral (1) to be
mathematically evaluated over the definite field-density range from
negative to positive infinity, using Fourier analysis.
The end result, deduced in (2.25) and generalized for various choices of
"measure" in (2.33), is the *mathematical* deduction that the evaluation
of (2) above over the definite range -oo to +oo is as follows:
Z=${-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) ] (2)
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 (3)
therefore moves through the path integration intact, and ends up inside
an impulse function in the space conjugate to spacetime. In this
formulation, the traceless EM field (L=.25 F^uv F_uv) clearly would
contribute, as would any trace matter kT=R.
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 (2). I am simply
addressing the *mathematical* problem of evaluating the definite
integral (1), and getting a real answer, which, according to what is
deduced here, the mathematics tells us is (2). 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 after several trials that were
errors, perhaps I finally have figured out the right way to do this.
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
Koobee Wublee
to each particular event. Who gives a damn about event A has
accumulated x amount of action, and event B has accumulated y amount
of action. By the way, the answer when integrating from -oo to +oo is
mostly likely infinite just like how much time would Voyager I and II
arrive at their respective destinations. Who cares since we don't
really know and have no way of tracking anymore. What is utmost
important is the density to this action. Under a special condition
where the integrating limits of this action are fixed, the results are
the Euler-Lagrange equations.
I am very certain that you do not understand the subject mattered. I
have shown your Professor Morin's work on the calculus of variations.
You apparently have not understood a single word of it.
Repeating the same nonsense over and over reminds me of your buddy,
Mr. Tucker, who after staring at Weinberg's book for over 40 years,
still has no idea what a Lagrangian is. You are in the same shoes.
Only fools treasure junk.
It is utterly amazing. Weinberg's book must cost big bucks while that
particular section of Professor Morin's description of Lagrangian
method is free over the internet. Professor Morin's is a must a
better place to study the very basics of GR.
<shrug>
Jay R. Yablon
the "measure problem" which Dr. Carlip raised in one a recent posts in a
parallel thread at
http://groups.google.com/group/sci.physics.research/browse_frm/thread/d15e35b971697965#.
I have therefore posted an update to this exercise at:
What the above now contains, just added, is a detailed discussion of how
to select the measure so that the overall path integral evaluates to
something that is invariant under general coordinate transformations. I
begin to introduce this discussion at equation (2.13), and the upshot is
that the only measure which meet this criteria of a generally-invariant
evaluation of the path integral is sqrt(-g), alone.
In brief, the "measure problem" is resolved by distinguishing between
local coordinate invariance and global coordinate invariance. The goal
is to end up with a result that is globally-invariant, even if the
variable of integration is itself not locally-invariant. One may think
of this approach as one of "hidden invariance." ***This all works,
because in taking a definite integral, even over a coordinate dependent
measure, the measure drops out from the overall expression that results
from evaluating the definite integral.***
In essence, the way this works is that when we do path integrals, we are
forced to select a particular system of coordinates, perform the
definite path integration in this system of coordinates, and obtain a
result. Then, we transform into a different system of coordinates,
again do the path integration, and obtain a second result. The goal is
to find that the second result is the same as the first result.
Repeating this ad infinitum, we choose system after system of
coordinates until we have exhausted every possible coordinate system,
and each time we do the integration, it is the goal to end up with the
same, invariant result. That is, no matter what system of coordinates
we choose, our path integral should yield the same invariant result.
Again, even though any particular choice of measure is a non-covariant
choice, the path integral should evaluate invariantly no matter what
coordinates we choose to represent the measure. This is then a
"globally" covariant result. This is what we mean by "hidden
covariance." This updated post demonstrates that sqrt(-g) is the only
measure which meets this goal and is the only truly invariant choice.
The final result of this effort, is in equation (3.19):
Z = $(-oo to +oo) Dsqrt(-g) exp[i S_EH] = delta^(4xoo)(L_EH) (1)
where the Einstein Hilbert action:
S_EH = $(-oo to +oo) sqrt(-g) L_EH d^4x (2)
and the Einstein Hilbert Lagrangian density is:
L_EH = (1/2 kappa) R + L_matter (3)
If you look closely at (3.19), and think about the Fourier
transformation:
$(-oo to +oo) dx exp[i x w] = delta (w) (4)
it should be self-evident that (1) (which is (3.19) in the linked file)
evaluates to a Dirac delta. In fact, in retrospect, one can omit just
about everything leading up to (3.19), stare at (3.19) closely for a few
moments, and realize that (3.19) is just a disguised form of (4), with a
coordinate-dependent measure that "washes out" no matter what coordinate
system one uses for the integration.
I just realized I am missing some sqrt(2pi) factors on the deltas. I'll
put this in a next draft, but those are non-essential to the main
result.
Jay.
Ken S. Tucker
On Nov 21, 9:42 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> Repeating the same nonsense over and over reminds me of your buddy,
> Mr. Tucker, who after staring at Weinberg's book for over 40 years,
> still has no idea what a Lagrangian is. You are in the same shoes.
> Only fools treasure junk.
>
> It is utterly amazing. Weinberg's book must cost big bucks while that
> particular section of Professor Morin's description of Lagrangian
> method is free over the internet. Professor Morin's is a must a
> better place to study the very basics of GR.
Playing the Lagrangian Card, is like playing the '10 of clubs'
out a '52 card deck'.
Can you parse the meaning of T_uv;w = 0 ?
If so I'll give you my 'Jack of Spades'
Ken
eric gisse
> Dear Miss Woblee...I see I have honorable mention.
>
> On Nov 21, 9:42 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>> Repeating the same nonsense over and over reminds me of your buddy,
>> Mr. Tucker, who after staring at Weinberg's book for over 40 years,
>> still has no idea what a Lagrangian is. You are in the same shoes.
>> Only fools treasure junk.
>>
>> It is utterly amazing. Weinberg's book must cost big bucks while that
>> particular section of Professor Morin's description of Lagrangian
>> method is free over the internet. Professor Morin's is a must a
>> better place to study the very basics of GR.
Except the mathematics that Morin's classical mechanics textbook contains is
still beyond wooby's grasp.
>
> Playing the Lagrangian Card, is like playing the '10 of clubs'
> out a '52 card deck'.
> Can you parse the meaning of T_uv;w = 0 ?
Really Ken, you don't know what that means?
Koobee Wublee
> Dear Miss Woblee...I see I have honorable mention.
Hmmm... Dear Mr. Tucker, you are still out of you fvcking mind --- as
usual.
> On Nov 21, 9:42 pm, Koobee Wublee wrote:
> > Repeating the same nonsense over and over reminds me of your buddy,
> > Mr. Tucker, who after staring at Weinberg's book for over 40 years,
> > still has no idea what a Lagrangian is. You are in the same shoes.
> > Only fools treasure junk.
>
> > It is utterly amazing. Weinberg's book must cost big bucks while that
> > particular section of Professor Morin's description of Lagrangian
> > method is free over the internet. Professor Morin's is a must a
> > better place to study the very basics of GR.
>
> Playing the Lagrangian Card, is like playing the '10 of clubs'
> out a '52 card deck'.
You should know by now I have no time to play with card games. I
guess you do. You should be spending time instead to study what a
Lagrangian is. <shrug>
Hint: Staring at a book for over 40 years without understanding what
a Lagragnian is should warrant you to move on. <shrug>
> Can you parse the meaning of T_uv;w = 0 ?
It is rather silly to do so since you have not specified what T_uv or
w is. It can just be anything imaginable. Apparently, you are totally
ignorant of the engineering world where specifications are indeed
bibles. <shrug>
You must be a salesman who knows nothing but pretends to know
everything. <shrug>
Salesmen are hopeless to understand anything. Self-styled physicists
can only cling on to their worlds of make-believes including the
fantasy parallel worlds. Only engineers have the very grasp on what
reality is. <shrug>
> If so I'll give you my 'Jack of Spades'
You can give me whatever you want. The fact remains that you have
stared at Weinberg's book for over 40 years and still have no fvcking
ideas of what a Lagrangian is. That is sad. Normally, this is very
laughable (at you), but I have only pity on you. Please get fvck
lost. You are beyond help due to your mental challenge as well as
your advancing age. <shrug>
Koobee Wublee
> Overnight, it became apparent to me that I should specifically discuss
> the "measure problem" which Dr. Carlip raised in one a recent posts in a
> parallel thread at
> http://groups.google.com/group/sci.physics.research/browse_frm/thread....
>
> I have therefore posted an update to this exercise at:
>
> http://jayryablon.files.wordpress.com/2009/11/fourier-path-integration-of-the-einstein-hilbert-action-41.pdf
>
> What the above now contains, just added, is a detailed discussion of how
> to select the measure so that the overall path integral evaluates to
> something that is invariant under general coordinate transformations. I
> begin to introduce this discussion at equation (2.13), and the upshot is
> that the only measure which meet this criteria of a generally-invariant
> evaluation of the path integral is sqrt(-g), alone.
This is total nonsense since g varies with the coordinate system that
you wish to describe the invariant geometry. For example, in
describing flat spacetime, you have the choice of using the linearly
rectangular coordinate system also known as the Euclidean coordinate
system and come up with the following 2-dimensional matrix known as
the metric in 3-dimensional space.
[g_e] = [1, 0, 0]
[0, 1, 0]
[0, 0, 1]
Or, you can also choose to use the common spherically symmetric polar
coordinate system in which the so-called metric becomes the following.
[g_p] = [1, 0, 0]
[0, r^2 cos^2(Latitude), 0]
[0, 0, r^2]
Notice [g_e] and [g_p] are drastically different in mathematics as
well as in concept. They are different metrics (2-dimensional
matrices), but they do describe the same geometry. In addition, their
determinants are very, very different. Now, you are beginning to
realize how the determinant of a matrix would affect the so-called
action. <shrug>
> In brief, the "measure problem" is resolved by distinguishing between
> local coordinate invariance and global coordinate invariance. The goal
> is to end up with a result that is globally-invariant, even if the
> variable of integration is itself not locally-invariant. One may think
> of this approach as one of "hidden invariance." ***This all works,
> because in taking a definite integral, even over a coordinate dependent
> measure, the measure drops out from the overall expression that results
> from evaluating the definite integral.***
It is all nonsense --- Einstein Dingleberry talk. GR remains a tool
to describe what the curvature of spacetime should be using any
observer's own choice of coordinate system. <shrug>
> In essence, the way this works is that when we do path integrals, we are
> forced to select a particular system of coordinates, perform the
> definite path integration in this system of coordinates, and obtain a
> result...
No. I have told you why already. <shrug>
> [Rest of non-scholarly works snipped]
You are mystifying yourself. <shrug>
Jay R. Yablon
news:2d9c5b74-ea11-473a...@f20g2000vbl.googlegroups.com...
You have missed my entire point.
Yes, you are completely correct about sqrt(-g) (and g_uv) being
dependent on the choice of coordinates. That is what makes the "measure
problem" a challenge, and it is unresolved today despite many papers
which people have written in the topic. BUT, when used as the measure
of integration, the sqrt(-g) drops out of the definite integral once it
is evaluated. The point is that I can put sqrt(-g) into one coordinate
system, say, the first (rectilinear) one you showed above, then do the
path integral calculation which is a definite integral, and get "result
1" which strips off the coordinate information encoded in sqrt(-g).
Then I can put sqrt(-g) into a second coordinate system, say, the second
(curvilinear) one you showed above, then do the path integral
calculation which is a definite integral, and get "result 2," which
strips off that coordinate information in sqrt(-g). All because
sqrt(-g) is the variable of integration.
But, that by itself is insufficient. It is also important that we get
the same answer each time, no matter which coordinate system we choose.
Certain choices of the measure do not achieve this. But, sqrt(-g) does.
That is, if we use sqrt(-g) as the measure of integration for the EH
action, then no matter what coordinate system we choose, we end up with
the exact same answer, which is an INVARIANT result. This is a form of
hidden symmetry -- that which is non-invariant on a local basis yields
an invariant result globally based on evaluating the definite integral.
I specifically discussed this starting at (2.14) and throughout section
3 of
http://jayryablon.files.wordpress.com/2009/11/fourier-path-integration-of-the-einstein-hilbert-action-41.pdf,
yet you act as if I ignored this. I suggest that you cease your
destructive practice of shooting first and asking questions later.
Jay
Ken S. Tucker
> On Nov 22, 5:41 pm, "Ken S. Tucker" wrote:
>
> > Dear Miss Woblee...I see I have honorable mention.
>
> Hmmm... Dear Mr. Tucker, you are still out of you fvcking mind --- as
> usual.
Miss Wobbly,
You cowardly choose to conceal your identity behind a
sexually rejected mystical Hindu female animal, please
don't take your bedroom problems out on us.
> > On Nov 21, 9:42 pm, Koobee Wublee wrote:
> > > Repeating the same nonsense over and over reminds me of your buddy,
> > > Mr. Tucker, who after staring at Weinberg's book for over 40 years,
> > > still has no idea what a Lagrangian is. You are in the same shoes.
> > > Only fools treasure junk.
>
> > > It is utterly amazing. Weinberg's book must cost big bucks while that
> > > particular section of Professor Morin's description of Lagrangian
> > > method is free over the internet. Professor Morin's is a must a
> > > better place to study the very basics of GR.
>
> > Playing the Lagrangian Card, is like playing the '10 of clubs'
> > out a '52 card deck'.
>
> You should know by now I have no time to play with card games. I
> guess you do. You should be spending time instead to study what a
> Lagrangian is. <shrug>
>
> Hint: Staring at a book for over 40 years without understanding what
> a Lagragnian is should warrant you to move on. <shrug>
>
> > Can you parse the meaning of T_uv;w = 0 ?
>
> It is rather silly to do so since you have not specified what T_uv or
> w is. It can just be anything imaginable. Apparently, you are totally
> ignorant of the engineering world where specifications are indeed
> bibles. <shrug>
Goto library, get Weinberg's "Grav&Cosmo" text, and
read something useful for a change.
> You must be a salesman who knows nothing but pretends to know
> everything. <shrug>
>
> Salesmen are hopeless to understand anything. Self-styled physicists
> can only cling on to their worlds of make-believes including the
> fantasy parallel worlds. Only engineers have the very grasp on what
> reality is. <shrug>
How about professional Sales Engineers?
> > If so I'll give you my 'Jack of Spades'
>
> You can give me whatever you want. The fact remains that you have
> stared at Weinberg's book for over 40 years and still have no fvcking
> ideas of what a Lagrangian is. That is sad. Normally, this is very
> laughable (at you), but I have only pity on you. Please get fvck
> lost. You are beyond help due to your mental challenge as well as
> your advancing age. <shrug>
Typical response from an old spinster.
Ken
Koobee Wublee
> "Koobee Wublee" wrote:
> > This is total nonsense since g varies with the coordinate system that
> > you wish to describe the invariant geometry. For example, in
> > describing flat spacetime, you have the choice of using the linearly
> > rectangular coordinate system also known as the Euclidean coordinate
> > system and come up with the following 2-dimensional matrix known as
> > the metric in 3-dimensional space.
>
> > [g_e] = [1, 0, 0]
> > [0, 1, 0]
> > [0, 0, 1]
>
> > Or, you can also choose to use the common spherically symmetric polar
> > coordinate system in which the so-called metric becomes the following.
>
> > [g_p] = [1, 0, 0]
> > [0, r^2 cos^2(Latitude), 0]
> > [0, 0, r^2]
>
> > Notice [g_e] and [g_p] are drastically different in mathematics as
> > well as in concept. They are different metrics (2-dimensional
> > matrices), but they do describe the same geometry. In addition, their
> > determinants are very, very different. Now, you are beginning to
> > realize how the determinant of a matrix would affect the so-called
> > action. <shrug>
>
> Yes, you are completely correct about sqrt(-g) (and g_uv) being
> dependent on the choice of coordinates.
That is not what the self-styled physicists are saying in the past 100
years or so. They are merely matrices, but the self-styled labeled
these matrices as tensors where they have mathemagical properties.
One of which is that they become coordinate independent. Just look
through all the old posts from these self-styled physicists. They all
are saying the metric g is a tensor which is independent of any choice
of coordinate system. Doesn’t your textbook say that? Why are you
going against what the self-styled physicists are saying?
> That is what makes the "measure
> problem" a challenge, and it is unresolved today despite many papers
> which people have written in the topic.
What problems are these exactly? You wish to describe the geometry.
In doing so, you choose a set of coordinate system to start with. The
metric then only applies to this chosen set of coordinate system in
describing the geometry. I don’t see any problems. It is merely a
limitation of mathematics. <shrug>
> BUT, when used as the measure
> of integration, the sqrt(-g) drops out of the definite integral once it
> is evaluated.
Every variables drop out of a definite integral. So, why is this sqrt
(-g) so special now?
> The point is that I can put sqrt(-g) into one coordinate
> system, say, the first (rectilinear) one you showed above, then do the
> path integral calculation which is a definite integral, and get "result
> 1" which strips off the coordinate information encoded in sqrt(-g).
OK, let’s call the result of this action to be E.
> Then I can put sqrt(-g) into a second coordinate system, say, the second
> (curvilinear) one you showed above, then do the path integral
> calculation which is a definite integral, and get "result 2," which
> strips off that coordinate information in sqrt(-g). All because
> sqrt(-g) is the variable of integration.
Let’s call the result of this action to be P.
> But, that by itself is insufficient.
What do you mean by itself? E and P are the results of the actions.
> It is also important that we get
> the same answer each time, no matter which coordinate system we choose.
But you don’t. <shrug>
I see where you are coming from. Take a closer look. For simplicity,
let’s leave out the Lagrangian of mass term.
** A = integral(R sqrt(- det(g)) dq^0 dq^1 dq^2 dq^3)
In the Eulcidean coordinate system,
** E = integral(R dt dx dy dz) / c
Where
** dq^0 = dt
** dq^1 = dx
** dq^2 = dy
** dq^3 = dz
** sqrt(- det(g)) = c
In the polar coordinate system,
** P = integral(R dt dr dLongitude dLatitude / (c r^2 |cos
(Latitude)|)
** dq^0 = dt
** dq^1 = dr
** dq^2 = dLongitude
** dq^3 = dLatitude
** sqrt(- det(g)) = c r^2 |cos(Latitude)|
Clearly, for your result to achieve this so-called symmetry, you must
manually modify the action to suit this purpose.
** A = integral(R sqrt(- det(g)) ds^0 ds^1 ds^2 ds^3)
Where
** ds^i = distance while dq^i = coordinate
See the difference? See the confusion on self-styled physicists’
part?
In the polar coordinate system,
** P = integral(R dt dr dLongitude dLatitude / c
Where
** ds^0 ds^1 ds^2 ds^3 = c r^2 cos(Longitude) dt dr dLongitude
dLatitude
The result to this action becomes the following.
** P = integral(R dt dr dLongitude dLatitude)
> Certain choices of the measure do not achieve this. But, sqrt(-g) does.
> That is, if we use sqrt(-g) as the measure of integration for the EH
> action, then no matter what coordinate system we choose, we end up with
> the exact same answer, which is an INVARIANT result.
E and P do not even have the same units.
> This is a form of hidden symmetry
Or merely a mathematical coincidence in which the mother nature is
full of. <shrug>
> -- that which is non-invariant on a local basis yields
> an invariant result globally based on evaluating the definite integral.
How can you can E and P to be the same if they do not even have the
same units? The fact remains that treating the metric as coordinate
independence will always come back to haunt you such as this case.
Why fight against the mother nature? Just accept the implications in
the mathematics and move on. <shrug>
Do you know what R is under the Euclidean coordinate system in flat
spacetime? I have not computed the polar coordinate system under flat
spacetime but would guess that it is also zero. For this example, why
go through such a great length to show zero = zero?
> I specifically discussed this starting at (2.14) and throughout section
> 3 of
> http://jayryablon.files.wordpress.com/2009/11/fourier-path-integratio...,
> yet you act as if I ignored this.
You are now behaving rather childish. I have no idea you have written
about that already. You have posted pieces of your works in piece
meals scattered in different newsgroups. I only monitor
sci.physics.relativity mostly, and I hate to read postings from
sci.physics.research because they don’t want any discussions from the
opposing camp. If your posting does not show up here in
sci.phsycis.relaivity with an interesting title, I just will not read
it. Don’t you think I have nothing else to do? You just cannot hold
that grudge that I have passed by your posting. <shrug>
Ken S. Tucker
On Nov 23, 11:16 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> On Nov 23, 5:40 am, "Jay R. Yablon" wrote:
The term "Relative Tensors" such as "sqrt(g)" having
weight W=+1, referred to archically as a tensor density,
it then follows the product of 'relative tensors' such as
W^-1 * W^+1 = 0 Weight and resolves that part to a
scalar (INVARIANT).
Relative tensor is a bit obscure but highly relevent to
to what Jay is balancing.
Regards
Ken S. Tucker