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Erik
Anmeldedatum: 28.03.2006
Beiträge: 565
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 Verfasst am: 08.07.2009, 15:51 Titel: |
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| Solkar hat Folgendes geschrieben: | @Erik:
| Zitat: | | Wrong. I think, it follows from (3.1) and (3.2) by integrating over y, dropping constant terms and equating the result with (3.4). |
First - you cannot "drop" constant terms on integration; if they occur in a product you can take take them from underneath the integral sign and "drag" them ahead of it as a multiplier.
You obviously missed differentiation and integration; if you had cast away named "lazyness" you would have detected your misconception yourself.
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Interesting, what is all so obvious to you.
No, I'm not talking about multiplicative constants, but additive constants arising by
integrating x-independent terms over y. I'm also not talking about these terms vanishing
in the process of integration, but dropping them afterwards. And the reason I can drop them is, because
they are irrelevant to the 4 dim. principle of least action (this dropping goes by the name "gauging
the action/lagrangian", AFAIK.) Randall and Sundrum also drop terms in eq (14) (resp. (15) in the arxiv version)
as indicated by the $ \supset $.
But if you don't like this you can carry them along and just ignore them.
| Zitat: |
Second - if you read the posting cited above again, you will find, that Randall-Sundrum(16) conceptually matches my second last step (before the division by $M_D^{2}$).
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Eq (16) "conceptually matches" my "integration over y". More clearly, that's exactly what it is.
But I don't see any resemblance to what you do. I even think the main point is missing in
your post (which is the substitution of the metric 3.2 into 3.1 and integration over y). The
division of the actions is an irrelevant step, that just lets you lose some generality. And you also don't
derive the result of this division.
| Zitat: |
Presumably this also would have been clear to you if you had cast away named "lazyness".
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I don't think that there's something unclear to me in your post. But you are the one, who
asked a question, implying that there's something you fail to understand. I just tried to
answer it.
As for me, I still don't see any major obstacles in deriving 3.5 in the G&M paper, which I
consider completely analog to (16) from Randall and Sundrum.
| Zitat: |
Generally speaking - I'm not going to make up for you being "lazy" (your own words) by teXing your approach.
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Fine, I don't demand anything like this. I simply don't consider it worth texing it by
myself. I think it's clear enough the way it is.
| Zitat: |
If you want continue this discussion, pls. do as I did and describe step-by-step thus equation-by-equation how you arrive at (3.5)
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My description is this: do the same as Randall and Sundrum did, step-by-step. What's unclear
with that? You just have to claculate the D-dim. R from D-dim. g, substitute it into the
D-dim. action and perform an integration over y. I believe (no, I don't know or
intend to prove) that this leads to 3.5.
BTW, you, too, did not derive 3.5., but simply outlined the derivation (at least in the
post you repeatedly referred to). Or am I missing something?
| Zitat: |
P.S.: Btw "teXing" such analysis is generally considered a courtesy to the readers. |
I don't care. I have made myself as clear as I think is necessary that has to be enough.
Moreover, no one's obliged to read my scribbling. |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 08.07.2009, 17:23 Titel: |
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About [1] Myers/Perry "Black Holes in Higher Dimensional Spacetimes", Princeton 1986:
I've eventually gotten access to this paper.
If G&M would be considered "complicated" [1] would be $e^{complicated}$ 
44 Pages with about 200 Equations (129 numbered); some of them relying on differential forms, some operating on Riemann(not "just" Ricci) -Tensor directly....
Much consideration of Reissner-Nordstroem and Kerr-solutions; the D-Dim Schwarzschild solution itself based on the paper [2] by Tangherlini.
I'll keep you updated.
Best regards,
Solkar
P.S.: @Erik:
I think I had been clear enough which efforts on your behalf would be needed for making me willing to continue this discussion with you.
| Solkar hat Folgendes geschrieben: |
If you want continue this discussion, pls. do as I did and describe step-by-step thus equation-by-equation how you arrive at (3.5)
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Erik
Anmeldedatum: 28.03.2006
Beiträge: 565
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| Verfasst am: 08.07.2009, 18:36 Titel: |
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| Solkar hat Folgendes geschrieben: |
P.S.: @Erik:
I think I had been clear enough which efforts on your behalf would be needed for making me willing to continue this discussion with you.
| Solkar hat Folgendes geschrieben: |
If you want continue this discussion, pls. do as I did and describe step-by-step thus equation-by-equation how you arrive at (3.5)
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LOL, yes, you want efforts you're quite obviously not willing to provide yourself. Resulting in me
doing all the work, which--judging by your participation in this thread--you're a hundred
times more interested in, than I am.
And what's wrong with the step-by-step description of Randall and Sundrum?
Sorry, but I'm not your algebra program and you're not dictating the terms of discussion. If
you want calculations, do them by yourself. If you're not interested in just how I think G&M
derived their result (instead of deriving it myself), just skip it. But anyway if you need excuse to
leave this discussion, feel free to use this one. |
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Barney
Anmeldedatum: 19.10.2008
Beiträge: 1538
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| Verfasst am: 08.07.2009, 18:56 Titel: |
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| Barney hat Folgendes geschrieben: |
that´s correct. (3.18) is an erroneous abbreviation. We get correct formulae from Wikipedia, too.
br |
Hi Solkar,
just for completeness, I have to correct my statement from above. (3.18) is true, but the corresponding text "and where $\Omega_{D-2}$ is the volume of the unit $\Omega_{D-2}$ sphere" (s. p. 13 first line) contains a typo. Correct would be "and where $\Omega_{D-2}$ is the surface of the unit $\Omega_{D-2}$ sphere".
Regards |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 08.07.2009, 19:11 Titel: |
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Hi Barney!
| Barney hat Folgendes geschrieben: | | Correct would be "and where $\Omega_{D-2}$ is the surface of the unit $\Omega_{D-2}$ sphere". |
For D=4 we got $4 \pi$; the $\Omega_{D-1}$ surface.
How do you get to $\Omega_{D-2}$?
Best Regards,
Solkar |
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ralfkannenberg
Anmeldedatum: 22.02.2006
Beiträge: 4788
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| Verfasst am: 08.07.2009, 19:15 Titel: |
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| Erik hat Folgendes geschrieben: | But anyway if you need excuse to
leave this discussion, feel free to use this one. |
Hi Leute,
jetzt bitte beruhigt Euch doch wieder und hört auf, Euch gegenseitig Kompetenzdefizite vorzuwerfen, von denen andere nur träumen würden, wenn sie diese auch nur annähernd erreichen könnten.
Das habt Ihr nun wirklich nicht nötig und ich bin überzeugt, dass Ihr diese unterschiedlichen Betrachtungen auch rein sachlich lösen könnt.
@Erik: Ich bin noch nicht dazu gekommen, Deine letzten Beiträge im Review-Thread zu beantworten, da ich schon früh einen Deiner Beiträge völlig missverstanden habe (kein Vorwurf, ganz klar mein Fehler !), d.h. ich muss alles zuerst nochmals lesen.
Freundliche Grüsse, Ralf |
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Barney
Anmeldedatum: 19.10.2008
Beiträge: 1538
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| Verfasst am: 08.07.2009, 19:57 Titel: |
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| Solkar hat Folgendes geschrieben: |
For D=4 we got $4 \pi$; the $\Omega_{D-1}$ surface.
How do you get to $\Omega_{D-2}$?
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Hi Solkar,
for D=4 we get $4 \pi$ and therefore the surface of a two-dimensional unit sphere. If one takes this sphere as a Riemanian manifold, one can easily identify the dimension as 2. Remember some usual [url=http://en.wikipedia.org/wiki/Atlas_(topology)]charts[/url] of this manifold.
It is just a definition of mathematicians  .
br |
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Erik
Anmeldedatum: 28.03.2006
Beiträge: 565
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| Verfasst am: 08.07.2009, 20:18 Titel: |
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| ralfkannenberg hat Folgendes geschrieben: | | Erik hat Folgendes geschrieben: | But anyway if you need excuse to
leave this discussion, feel free to use this one. |
Hi Leute,
jetzt bitte beruhigt Euch doch wieder und hört auf, Euch gegenseitig Kompetenzdefizite vorzuwerfen, von denen andere nur träumen würden, wenn sie diese auch nur annähernd erreichen könnten.
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Hallo Ralf,
woher hast du diesen Vorwurf? Aus meinen Beiträgen jedenfalls nicht.
Ich möchte festhalten, niemandem etwas vorgeworfen zu haben, schon gar nicht mangelnde
Kompetenz.
(Ich kann mich übrigens auch nicht an solch einen Vorwurf seitens Solkar erinnern.
Naja, vielleicht die dauernd unterstellten "misconceptions", die angeblich noch durch
ungenügendes Lesen seiner Beiträge verschuldet sein sollten. Sowas nehme ich aber
nicht ernst.)
| Zitat: |
Das habt Ihr nun wirklich nicht nötig und ich bin überzeugt, dass Ihr diese unterschiedlichen Betrachtungen auch rein sachlich lösen könnt.
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Dieser letzte Versuch eines sachlichen Beitrages wurde leider mit fadenscheinigen Gründen ignoriert. (Mich würde auch sonst interessieren,
was du unsachlich fandest.)
| Zitat: |
@Erik: Ich bin noch nicht dazu gekommen, Deine letzten Beiträge im Review-Thread zu beantworten, da ich schon früh einen Deiner Beiträge völlig missverstanden habe (kein Vorwurf, ganz klar mein Fehler !), d.h. ich muss alles zuerst nochmals lesen.
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Kein Problem, mach dir nicht immer zu viel Arbeit und gib dir dann auch noch die Schuld für
Mißverständnisse. Das ist nicht nötig, passiert halt. |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 09.07.2009, 02:04 Titel: |
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| Barney hat Folgendes geschrieben: | | two-dimensional unit sphere. If one takes this sphere as a Riemanian manifold, one can easily identify the dimension as 2. |
Thank you Barney, that was a good one!
(And important for my grasp on Diff'geo - somehow my neurons had not linked that to this topic...  )
Thank you very much!
Best Regards,
Solkar
P:S.: But to complete the "knot" [1]Myers&Perry use $d\Omega_{N-1}^2$ in their eq.(2.1) a N(?)-dim Schwarzschild's N-2(?) sub-manifold (all except "r" and "t", as usual)
BUT
just in the paragraph below they explain it by stating $d\Omega_{N}^2$ was "the line element of the unit N-sphere".
No kidding!
P.P.S: It's quite a pity that [1] is not freely available, it's imho a brilliant piece of higher maths! |
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Barney
Anmeldedatum: 19.10.2008
Beiträge: 1538
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| Verfasst am: 10.07.2009, 19:36 Titel: |
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Hi Solkar, Hi all,
there is a possibility to derive (3.5) from the G&M-paper under certain assumptions. For a constant warp factor, i.e. A = const. (a real number), the Ricci scalar is simply a sum
$R_D = R_4(x)\quad + \quad R(y)$. The determinant -g_D is a composition of three multiplicative terms, as written the post above. Now, together with the restriction R(y) = 0, (3.5) becomes valid.
The general case is in my opinion much more complex. If the warp factor is a function of the compact variables y, we have no linearity for the Ricci scalar.
br |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 10.07.2009, 20:07 Titel: |
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| Barney hat Folgendes geschrieben: | there is a possibility to derive (3.5) from the G&M-paper under certain assumptions. For a constant warp factor, i.e. A = const. (a real number), the Ricci scalar is simply a sum.
$R_D = R_4(x)\quad + \quad R(y)$. The determinant -g_D is a composition of three multiplicative terms, as written the post above. Now, together with the restriction R(y) = 0, (3.5) |
Precisely,
| Solkar hat Folgendes geschrieben: |
wichtig ist dabei, dass ein hier
| Erik hat Folgendes geschrieben: | \[
R(x,y)\sqrt{-g}= \sqrt{g_{D-4}}e^{2A}\ R_4(x)\sqrt{-g_4} + R'(y)
\] |
R'(y) genannter Skalar eben verschwinden muss um zu (3.5) gelangen zu können; |
if R'(y) vanishes it becomes easier.
But we would still need that named "division" to finally get rid of the g_4 factor and the R_4 scalar.
| Barney hat Folgendes geschrieben: |
The general case, where the warp factor is a function of the compact variables y, is in my opinion much more complex, because the linearity doesn´t hold in this case.
br |
Then one can factorize the new D-4 Ricci this way, under the premises of R&S for all D that would do for (3.5)
Best regards,
Solkar |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 10.07.2009, 21:52 Titel: |
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Addendum:
It will imo become interesting to have close look at the partial fraction decomposition of the integral fraction; regularly we could remove the Singularity in R by cancelling out the fractions top and down , but we might find some obstacles on the way there; especially if any physical quantities would lose their meaning on the way. |
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Barney
Anmeldedatum: 19.10.2008
Beiträge: 1538
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| Verfasst am: 10.07.2009, 23:22 Titel: |
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| Barney hat Folgendes geschrieben: | the Ricci scalar is simply a sum
$R_D = R_4(x)\quad + \quad R(y)$
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erratum:
just for completeness. The correct formula is $\mathcal{R} = e^{-2A}\mathcal{R}_4(x)\quad + \quad \mathcal{R}(g_{mn}(y))$. |
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Erik
Anmeldedatum: 28.03.2006
Beiträge: 565
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| Verfasst am: 10.07.2009, 23:27 Titel: |
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| Solkar hat Folgendes geschrieben: | | Barney hat Folgendes geschrieben: | there is a possibility to derive (3.5) from the G&M-paper under certain assumptions. For a constant warp factor, i.e. A = const. (a real number), the Ricci scalar is simply a sum.
$R_D = R_4(x)\quad + \quad R(y)$. The determinant -g_D is a composition of three multiplicative terms, as written the post above. Now, together with the restriction R(y) = 0, (3.5) |
Precisely,
| Solkar hat Folgendes geschrieben: |
wichtig ist dabei, dass ein hier
| Erik hat Folgendes geschrieben: | \[
R(x,y)\sqrt{-g}= \sqrt{g_{D-4}}e^{2A}\ R_4(x)\sqrt{-g_4} + R'(y)
\] |
R'(y) genannter Skalar eben verschwinden muss um zu (3.5) gelangen zu können; |
if R'(y) vanishes it becomes easier.
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No, if you have the above form, everything is done.
And I think the proof is complete now without any loss of generality like constant A.
It's very easy (if I didn't overlook something important) and also avoids any extensive
Ricci calculus, but uses only ideas that have already come up in this thread. Here it is:
The metric
\[ g = e^{2A(y)}g_{\mu\nu}(x) dx^\mu dx^\nu + g_{mn}(y)dy^n dy^m \]
is conformally equivalent to a metric of the form
\[ \bar g = g_{\mu\nu}(x)dx^\mu dx^\nu + \bar g_{mn}(y) dy^n dy^m. \]
the conformal factor being $ e^{A} $. But the curvature scalar of this $ \bar g $ actually is of
the form $ R_4(x) + \bar R(y) $. (With $ R_4 $ calculated solely from
$ g_{\mu\nu} $. The curvature forms are completely seperated
for each submanifold.)
So we have (by a known theorem, cf. eq. (D.9) of Wald, General Relativity) for the scalar R
of original metric, in extenso
\[
R = e^{-2A(y)}\left[ R_4(x) + \bar R(y) - 4(D-1)g^{mn}\nabla_m\nabla_n A- 4(D-2)(D-1)g^{mn}\nabla_m A\nabla_n A\right]
\]
Everything in parenthesis save $ R_4 $ is x-independent.
Also the original metric is block diagonal, so we have
\[ \sqrt{-g_D} = e^{4A}\ \sqrt{-g_4}\sqrt{g_{D-4}} \]
Putting everything together, we arrive at the desired
\[ R\sqrt{-g_D} = e^{2A}\sqrt{g_{D-4}}\ \sqrt{-g_4} R_4(x) + R'(y)\sqrt{-g_4}. \]
(there is an additional factor $ \sqrt{-g_4} $ compared to the above formula, but this does no harm
and the original decomposition of R can be generalized in this respect.)
So, the D-dimensional action now is by definition
\[ S_D[g] =M^{D-2}_D \int dx dy \sqrt{-|g_D|} {R\over 2}
= M^{D-2}_D\left(\int dy e^{2A}\sqrt{g_{D-4}} \int dx \sqrt{-g_4} {R_4(x)\over 2} + \int dx \sqrt {-g_4}\int dy R'(y)\right)
\]
| Zitat: |
But we would still need that named "division" to finally get rid of the g_4 factor and the
R_4 scalar.
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No, nothing like this. From now on we can proceed like Randall and Sundrum. Integration of
R'(y) over y gives a term that will give a constant contribution $ \sim V_4 $, the
4-dimensional volume, to the 4 dim. action. That contribution may be infinite, but the
precise value of it is of no further consequence. The remaining terms just constitute
the ordinary 4-dimensional action.
\[ S_4 \equiv M^2_4\int dx {R_4\over 2}\sqrt{-g_4} = M^{D-2}_D \int dy e^{2A}\sqrt{g_{D-4}} \int dx \sqrt{-g_4} {R_4(x)\over 2} + O(V_4) \]
In order for this to hold for every metric $ g_4 $ it is necessary that
\[ M^2_4 = M_D^{D-2} \int dy e^{2A}\sqrt{g_{D-4}}. \qquad (3.5) \]
(The missing factors of $ 2\pi $ come from a different definition of the D-dimensional Planck mass.
Use the substitution $ M_D^{D-2} \rightarrow {M_D^{D-2} \over (2\pi)^{D-4}} $,
cf. (3.3) and (3.1) in G&M and compare with my D-dim. action.)
Dividing by something that is not guaranteed to be nonzero does no good, but it's unnecessary
here .
(Think about my example $ ax^2+ bx = x $. In order to conclude b=1, a=0 you need not/must not
divide by x, but just assume, that the equation holds for every x.) |
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Solkar
Anmeldedatum: 29.05.2009
Beiträge: 293
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| Verfasst am: 10.07.2009, 23:51 Titel: |
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Gentlemen,
to discuss possible R-Singularities/Poles (not to be missed with the famous Schwarzschild-Singularities/Poles) the long form of this
| Solkar hat Folgendes geschrieben: | Man setzt den Vorfaktor des ersten Integrals von (3.1) durch (3.3) und teilt den Integranden von in (3.1) durch $(2\pi)^{D-4}$
Dann setzt man $S^{grav}_D = S_4$; stellt die Gleichungen $S_4$ und $S^{grav}_D$ nach den Termen in $M$ um, setzt den Quotienten an, "dividiert" die Integrale und bringt $M^{D-2}_D$ nach rechts.
Dann teilt man durch $M^{2}_D$.
Dann bleibt rechts vor dem Integral von (3.5) $M^{D-4}_D$ übrig. |
(its original context was here) could imo be of assistance.
Here we go:
---
$S_D=\underbrace{\frac{1}{ 8\pi G_D} \int d^D x \sqrt{-g_D} ~\frac{1}{2} \cal{R}_D}_{S_D^{grav}} + \int d^D x \sqrt{-g_D} {\cal L} \quad$ (G&M eq(3.1) re-indexed) (VII.1)
Now fitting in D-dim Planck mass from G&M eq(3.3)...
$S_D^{grav }= ~\frac{M_D^{D-2}}{2} \int \frac{d^Dx}{{(2\pi)}^{D-4}} \sqrt{-g_D} \mathcal{R}_D$(VII.2)
The same procedure on a Minkowskian Einstein-Hilbert action yields this:
$S_4^{grav}= ~\frac{M_4^2}{2} \int d^4x \sqrt{-g_4} \mathcal{R}_4 \quad$(G&M eq(3.4)) (VII.3)
After we eliminated unknown $G_D$ here's is a normalization of the Integrals.
Pls kindly attend that the action integrals of Hamilton's Principle <![CDATA[http://en.wikipedia.org/wiki/Hamilton's_principle]]>(*) are meaningful in terms of stationary points on the Lagrangian function space, not because of their absolute value (and integrals are linear functionals).
$S_4^{grav} = S_D^{grav }$(VII.4)
Now some basic algebra:
$\Rightarrow \frac{M_4^2}{2} \int d^4 x \sqrt{-g_4} \mathcal{R}_4 = \frac{M_D^{D-2}}{2} \int \frac{d^Dx}{{(2\pi)}^{D-4}} \sqrt{-g_D} \mathcal{R}_D$(VII.5)
$\Rightarrow \frac{M_4^2}{M_D^{D-2}} = \frac{\int \frac{d^Dx}{{(2\pi)}^{D-4}} \sqrt{-g_D} \mathcal{R}_D}{\int d^4 x \sqrt{-g_4} \mathcal{R}_4}$(VII.6)
Now Fubini-Decomposition, replacing of $R_D$ with $R_4$ as shown at R&S eq(15) and applying the conformant re-scaling sampled here
$\Rightarrow \frac{M_4^2}{M_D^{D-2}} = \frac{\int d^4x \int \frac{d^{D-4}x}{{(2\pi)}^{D-4}}\sqrt{g_{D-4}} e^{2A} \sqrt{-g_4} \mathcal{R}_4} {\int d^4x \sqrt{-g_{4}}\mathcal{R}_4}$(VII.7)
$R_4$ and $g_4$ are invariant on non-Minkowskian coordinates
$\Rightarrow \frac{M_4^2}{M_D^{D-2}} = \frac{\int d^4x \sqrt{-g_4} \mathcal{R}_4 \int \frac{d^{D-4}x}{{(2\pi)}^{D-4}}\sqrt{g_{D-4}} e^{2A}} {\int d^4x \sqrt{-g_{4}}\mathcal{R}_4}$(VII.8)
The rest is basic algebra
$\Rightarrow \frac{M_4^2}{M_D^{D-2}} = \int \frac{d^{D-4}x}{{(2\pi)}^{D-4}}\sqrt{g_{D-4}} e^{2A}$(VII.9)
$\Rightarrow M_4^2 = M_D^{D-2} \int \frac{d^{D-4}x}{{(2\pi)}^{D-4}}\sqrt{g_{D-4}} e^{2A}$ (VII.10)
$\Rightarrow \underline{\frac{M_4^2}{M_D^{2}} = {M_D^{D-4}} \int \frac{d^{D-4}x}{{(2\pi)}^{D-4}}\sqrt{g_{D-4}} e^{2A}}$.(G&M eq(3.5) left) (VII.11)
(*) Does anyone know how to mask a " ' " in BBCode?[/url] |
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