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Barney
Anmeldedatum: 19.10.2008 Beiträge: 1538
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Verfasst am: 03.07.2009, 21:37 Titel: |
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ralfkannenberg hat Folgendes geschrieben: |
He told me of some current gravitational experiments in the micrometer-range most likely still behaving keplerian and this would if confirmed exclude the case of 2 Extra Dimensions - fitting well to you D=6 and he told me that they hope to have results for the nanometer-Range until 2015 excluding the case of 3 extra-dimensions, fitting well to you case D=7.
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btw: it´s not forbidden to separate such sentences into two or more appropriate sentences . Nevertheless, thanx for the additional infos. |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 04.07.2009, 02:48 Titel: |
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Hi!
About G&M eq.(3.15) to eq.(3.18):
I think it'd be useful to analyze those equations in reverse order
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Starting with (3.18):
The $\Gamma$-Function in the Denominator is basically a real (even complex) continuation of the natural factorial; e.g
$\Gamma(n) = (n-1)!$ (III.1)
for $n \epsilon \mathbb{N}\setminus \{0\}$
This yields the Denominator of (3.18) for even $(D-1)$, and the famous
$\Gamma(\frac{1}{2}) = \sqrt{\pi}$ (III.2)
together with
$\Gamma(z+1) = z \cdot \Gamma(z)$ (III.3) (*),
will yield $\Gamma$ for odd $(D-1)$ in (3.18)
Thus we get
$\Gamma(\frac{1}{2}) = \sqrt{\pi}$ (III.4)
$\Gamma(1) = 0! \equiv 1$ (III.5)
$\Gamma(\frac{3}{2}) = \Gamma(\frac{1}{2} + 1) = \frac{1}{2}\Gamma(\frac{1}{2}) = \frac{1}{2}\sqrt{\pi}$ (III.6)
$\Gamma(2) = 1! = 1$ (III.7)
$\Gamma(\frac{5}{2}) = \Gamma(\frac{3}{2} + 1) = \frac{3}{2}\Gamma(\frac{3}{2}) = \frac{3}{4}\sqrt{\pi}$ (III.8 )
...
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When we use this for (3.18) we get
For$D=4$
$ \Omega_{4-2} = \frac{2 \pi^{(4-1)/2}}{\Gamma((4-1)/2)}=\frac{2 \pi^{3/2}}{\frac{1}{2} \sqrt{\pi}} = 4 \sqrt{\frac{\pi^3}{\pi}} = 4 \pi $ (III.9)
For$D=5$
$ \Omega_{5-2} = \frac{2 \pi^{(5-1)/2}}{\Gamma((5-1)/2)}=\frac{2 \pi^{4/2}}{1} = 2 \pi^2 $ (III.10)
Both are apparently the D-1 unit sphere surface areas, so G&M's term "volume of the unit D-2 sphere" (above eq.(3.18), atop their p.13; PDF p.14 respectively) is imo at least a little misleading.
What do you think?
Best regards,
Solkar
(*) e.g. here:
Code: | Fischer/Lieb "Funktionentheorie"
Komplexe Analysis in einer Veränderlichen
9.Auflage 2008
Vieweg, Wiesbaden, Germany, 2008
| at p.205 used as a premise for a function $f$ which shall have $\Gamma$-behaviour |
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Barney
Anmeldedatum: 19.10.2008 Beiträge: 1538
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Verfasst am: 04.07.2009, 08:05 Titel: |
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Solkar hat Folgendes geschrieben: |
What do you think?
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Hi Solkar,
that´s correct. (3.18) is an erroneous abbreviation. We get correct formulae from Wikipedia, too.
br |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 04.07.2009, 13:22 Titel: |
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Hi Barney!
A wild guess of mine:
General Stokes' Theorem
$\int \limits_{V}^{} d\omega = \oint \limits_{\partial V} \omega $ (IV.1)
resp one of its consequences
$\int \limits_{V}^{} \langle \vec{\nabla} , \vec{\Psi} \rangle dV = \oint \limits_{\partial V} \vec{\Psi} d\vec{\Sigma} $ (IV.2) (Gauss' Theorem, e.g Maxwell I)
or - even less likely
$\int \limits_{V}^{} \vec{\nabla} \times \vec{\Psi} d\vec{\Sigma} = \oint \limits_{\partial V} \vec{\Psi} d\vec{r} $ (IV.3) (Classical Stokes' Theorem, e.g. Maxwell-Faraday)
applied in any tricky way on Einstein's "space-field" (here labelled $\Psi$ by me)
It would still not be a Lebesgue-Volume, but maybe it has some hidden ART-(or ST-)meaning, which does not disclose itself here.
Greetings,
Solkar |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 04.07.2009, 13:50 Titel: |
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At least with
$D=4$
$ \Omega_{4-2} = 4 \pi $ (from III.9)
we might get for (3.17)
$k_4 = \frac{2{(2 \pi)}^{(4-4)}}{(4-2) 4 \pi} = \frac{1}{4 \pi} $ (V.1)
thus for (3.16) expressed in NU:
$ R_4(M) = {\frac{1}{\frac{1}{sqrt{8 \pi}}} \lbrace { \frac{ \frac{1}{4 \pi}M}{\frac{1}{sqrt{8 \pi}} } \rbrace^{1/(4-3)} = \frac{{(sqrt{8 \pi})}^2}{4 \pi} M = 2M [NU] \underbrace{\equiv}_{? well, maybe} \frac{2 G M }{c^2} [SI]$ (V.2)
There are samples when NU don't provide much clarity after all.... |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 07.07.2009, 16:49 Titel: |
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About
$ds^2 = -\left[ 1- \left({R(M)\over r}\right)^{D-3}\right] dt^2 +{1\over 1- \left(\frac{R(M)}{r}\right)^{D-3}}dr^2 + r^2 d\Omega^2$ (G&M eq.(3.15))
This is said to be a D-dimensional Schwarzschild solution ("SE") ; and obviously it formally yields the well-known 4-dimenssional Schwarzschild metric for D=4.
Question is, what validity this equation will have for higher dimensions D>4.
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G&M do not directly cite a source for this generalization, but they do mention
Myers, R.C., and Perry, M.J., “Black Holes In Higher Dimensional Space-Times”, Ann. Phys. (N.Y.), 172, 304, (1986). (their ref [29], my ref [1] resp "M&P" subsequently)
with reference to Kerr-solutions (and the end of their Chapter 3.2.1).
Unfortunately, neither [1] nor
Tangherlini, F.R., “Schwarzschild field in n dimensions and the dimensionality of space problem”, Nuovo Cimento, 27, 636, (1963). ([2] or "S/T" subsequently)
("S/T" because "Schwarzschild-Tangherlini-" model/solution/equation seems to be a usual term in higher-dim ART)
are freely available.
I'm working on getting access to these papers.
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There are, however, some interesting free sources covering the topic
"Black Holes in Higher Dimensions"
Roberto Emparan and Harvey S. Reall
http://www.livingreviews.org/lrr-2008-6
([3] resp "E&R" subsequently)
and
Teo, E., “Black diholes in five dimensions”, Phys. Rev. D, 68, 084003, (2003). Related online version (cited on 14 February 2008):
http://arxiv.org/abs/hep-th/0307188
([4] resp "arXiv:hep-th/0307188v2" resp "T-BDH" subsequently)
===
It is, however, somewhat tempting trying to verify G&M (3.15) by own efforts.
A)Complexification
==================
Given the standard form
$ds^2 = c^2 B(r) dt^2 - A(r) dr^2 - r^2 (d\theta + sin^2\theta d\phi^2) $ (VI.2)
used for deduction of 4-dim SE
Given A and B were (real-valued) components of a complex function
$f(z) = A(|z|) + i B(|z|) $ (VI.3)
one might get some useful features for A's and B's existence and uniqueness from that; and
$f(r+it) = A(r,t) + i B(r,t) $ (VI.4)
could be a good approch to get to non-stationary solutions (e.g. Kerr)
But that is just a guess of mine.
B)Induction over $D\varepsilon\mathbb{N}$ and Gaussian Elimination
========================================
Let the basis be SE for 4 dim.
For the step:
With
$R_{\mu\nu} = R^{\alpha}_{\mu\nu\alpha}=\frac{\partial{\Gamma^{\alpha}_{\mu\alpha}}}{\partial x^{\nu}} - \frac{\partial{\Gamma^{\alpha}_{\mu\nu}}}{\partial x^{\alpha}} + \Gamma^{\alpha}_{\beta\nu}\Gamma^{\beta}_{\alpha\mu}-\Gamma^{\alpha}_{\beta\alpha}\Gamma^{\beta}_{\mu\nu}$ (VI.5)
and
$\Gamma^{\beta}_{\mu\nu}=\frac{1}{2}g^{\alpha\beta} \left(\frac{\partial g_{\alpha\nu}}{\partial x^{\mu}} + \frac{\partial g_{\mu\alpha}}{\partial x^{\nu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}} \right) $ (VI.6)
On could do away with plenty of the terms by either the steps's premise ("valid for D-1")
and the classical SE premises SE-metric being
- vacuum
- spherical
- stationary
That would presumably not be not very difficult, but very "lengthy"; maybe MAPLE or another high-end CAS would be helpful; "Maxima" (which I use) does not seem overly appropriate for supporting this task
===============
In any case, it would be efficient to wait for papers [1] and [2] to become available; otherwise we could easily "reinvent the wheel".
===============
There is, however, something we can discuss right now:
SE and Kerr are both "vacuum solutions", meaning that the stress-energy tensor $T_{\mu\nu}$vanishes.
For G&M (3.1)-(3.5) $T_{\mu\nu}$ is not "allowed" to vanish, because
With $T_{\mu\nu}=0$ we'd get
$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = \pm \kappa T_{\mu\nu} = 0$ (VI.7)
$\Rightarrow g^{\mu\nu} R_{\mu\nu} - \frac{1}{2} g^{\mu\nu} R g_{\mu\nu} = g^{\mu\nu} \quad \cdot 0 $
$\Rightarrow R - \frac{1}{2} R \cdot 4 = 0 $
thus
$R = 0 $ (VI.8)
which would make the Integrals of G&M(3.1)-(3.4) useless.
What's the justification for G&M deploying SE/Kerr in 3.2.1 nevertheless?
Best regards,
Solkar |
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Erik
Anmeldedatum: 28.03.2006 Beiträge: 565
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Verfasst am: 07.07.2009, 19:13 Titel: |
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Solkar hat Folgendes geschrieben: |
$f(r+it) = A(r,t) + i B(r,t) $ (VI.4)
could be a good approch to get to non-stationary solutions (e.g. Kerr)
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The Kerr-solution is stationary, BTW.
Zitat: |
SE and Kerr are both "vacuum solutions", meaning that the stress-energy tensor $T_{\mu\nu}$vanishes.
For G&M (3.1)-(3.5) $T_{\mu\nu}$ is not "allowed" to vanish, because
With $T_{\mu\nu}=0$ we'd get
$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = \pm \kappa T_{\mu\nu} = 0$ (VI.7)
$\Rightarrow g^{\mu\nu} R_{\mu\nu} - \frac{1}{2} g^{\mu\nu} R g_{\mu\nu} = g^{\mu\nu} \quad \cdot 0 $
$\Rightarrow R - \frac{1}{2} R \cdot 4 = 0 $
thus
$R = 0 $ (VI.8)
which would make the Integrals of G&M(3.1)-(3.4) useless.
What's the justification for G&M deploying SE/Kerr in 3.2.1 nevertheless?
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Equations 3.1, 3.4 have to hold for every metric of the form 3.2
(whether it's a solution to Einstein's equations or not) with constant M.
They just define the Einstein-Hilbert-action. So the vanishing of the action
for some solutions (vacuum) is simply of no consequence for 3.5 and is, of course allowed.
Put another way: S[g]=0 is not "useless", but simply a constraint on g. |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 07.07.2009, 19:53 Titel: |
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Erik hat Folgendes geschrieben: | /b]
Equations 3.1, 3.4 have to hold for every metric of the form 3.2
(whether it's a solution to Einstein's equations or not) with constant M.
They just define the Einstein-Hilbert-action. So the vanishing of the action
for some solutions (vacuum) is simply of no consequence for 3.5 and is, of course allowed.
Put another way: S[g]=0 is not "useless", but simply a constraint on g. |
In (3.1)-(3.5) a D-dim Planck mass is defined in Terms of the gravitational part of the EH-action.
If the respective integrals vanished there would be nothing defined.
Best regards,
Solkar |
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Erik
Anmeldedatum: 28.03.2006 Beiträge: 565
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Verfasst am: 07.07.2009, 20:05 Titel: |
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Solkar hat Folgendes geschrieben: | Erik hat Folgendes geschrieben: | /b]
Equations 3.1, 3.4 have to hold for every metric of the form 3.2
(whether it's a solution to Einstein's equations or not) with constant M.
They just define the Einstein-Hilbert-action. So the vanishing of the action
for some solutions (vacuum) is simply of no consequence for 3.5 and is, of course allowed.
Put another way: S[g]=0 is not "useless", but simply a constraint on g. |
In (3.1)-(3.5) a D-dim Planck mass is defined in Terms of the gravitational part of the EH-action.
If the respective integrals vanished there would be nothing defined.
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No. eqs. (3.1)-(3.5) do not define the planck masses in terms of anything, but establish a relation between D- and 4-dimensional
Planck masses (3.5) by defining the warped volume. Neither the Ms nor the relations between them depend on the value of S.
Maybe your error is to assume, that one calculates one M for every solution g of Einsteins equation? |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 07.07.2009, 21:11 Titel: |
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Erik hat Folgendes geschrieben: | No. eqs. (3.1)-(3.5) do not define the planck masses in terms of anything, but establish a relation between D- and 4-dimensional Planck masses (3.5) by defining the warped volume |
Erik, your arguments lack logic.
Although we've left this point since quite a while - I'm under the impression that you - still - don't take eq.(3.5) into account.
Whatever one does with (3.1)-(3.4) at the end the whole (3.5) must be the result.
Erik hat Folgendes geschrieben: | Maybe your error is to assume, that one calculates one M for every solution g of Einsteins equation? |
Anyway - you've yet shown no clear alternate concept to get to eq.(3.5).
I think you'd agree that a division by zero is to to be avoided or don't you?
Best Regards,
Solkar |
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Erik
Anmeldedatum: 28.03.2006 Beiträge: 565
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Verfasst am: 07.07.2009, 21:35 Titel: |
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Solkar hat Folgendes geschrieben: | Erik hat Folgendes geschrieben: | No. eqs. (3.1)-(3.5) do not define the planck masses in terms of anything, but establish a relation between D- and 4-dimensional Planck masses (3.5) by defining the warped volume |
Erik, your arguments lack logic.
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And this comment lacks arguments.
Zitat: |
Although we've left this point since quite a while - I'm under the impression that you - still - don't take eq.(3.5) into account.
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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). But I'm too lazy to check by direct
calculation.
Zitat: |
Whatever one does with (3.1)-(3.4) at the end the whole (3.5) must be the result.
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Certainly not. Take this example: If the equation
ax+bx² = x has to hold for every x (as 3.1,3.4 hold for every metric 3.2), one can infer that a=1, b=0.
But if I substitute x=0, I can infer nothing. Substituting the vacuum solution into the action, as you
do, is pointless. The planck masses do not depend on a specific solution.
Zitat: |
Erik hat Folgendes geschrieben: | Maybe your error is to assume, that one calculates one M for every solution g of Einsteins equation? |
Anyway - you've yet shown no clear alternate concept to get to eq.(3.5).
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I sketched it above. What's unclear about that?
Zitat: |
I think you'd agree that a division by zero is to to be avoided or don't you?
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You still think, one gets to 3.5 by devision of action integrals? That's surely wrong. |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 07.07.2009, 21:59 Titel: |
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Erik hat Folgendes geschrieben: | I sketched it above. |
On the contrary, Erik, I have "sketched" a way to get to (3.5); pls kindly attend the lower third of this posting of mine. You yet failed to do similar, and this
Erik hat Folgendes geschrieben: | But I'm too lazy to check by direct calculation. |
is most likely the source of your misconceptions.
(Sorry to say this, but "lazyness" is hardly a source of understanding)
Erik hat Folgendes geschrieben: | You still think, one gets to 3.5 by devision of action integrals? That's surely wrong. |
On the contrary, that is exactly the way to get to (3.5). Pls kindly attend that there is a division involved to get rid of the 4-dim part; I've already explained this thoroughly at the posting cited above.
Best regards,
Solkar |
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Erik
Anmeldedatum: 28.03.2006 Beiträge: 565
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Verfasst am: 07.07.2009, 22:19 Titel: |
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Solkar hat Folgendes geschrieben: | Erik hat Folgendes geschrieben: | I sketched it above. |
On the contrary, Erik, I have "sketched" a way to get to (3.5); pls kindly attend the lower third of this posting of mine and you yet failed to do similar,
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So we both sketched one way. You, for whatever reason, prefer to ignore mine. Your problem.
(And, BTW, I already read your post and think you're wrong, that's all.)
Zitat: |
Erik hat Folgendes geschrieben: | But I'm too lazy to check by direct calculation. |
is most likely the source of your misconceptions.
(Sorry to say this, but "lazyness" is hardly a source of understanding)
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So have you checked that 3.1,3.2 don't lead to 3.5 the way I described? If so I would be interested. Otherwise I don't
believe in any misconceptions of mine. But, indeed I have no ambition to understand every technical detail of the paper.
That's your business, as it seems.
Zitat: |
Erik hat Folgendes geschrieben: | You still think, one gets to 3.5 by devision of action integrals? That's surely wrong. |
On the contrary, that is exactly the way to get to (3.5). Pls kindly attend that there is a division involved to get rid of the 4-dim part; I've already explained this thoroughly at the posting cited above.
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And as you can see, the way I described above, does not involve any such division. (And, also, I thought
your thorough explanation still has some unclear points, too. But even if you get to 3.5 by your
division, you're still not forced to insert the vacuum solution and divide by zero.) |
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Erik
Anmeldedatum: 28.03.2006 Beiträge: 565
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Verfasst am: 08.07.2009, 10:59 Titel: |
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Erik hat Folgendes geschrieben: | Solkar hat Folgendes geschrieben: |
Although we've left this point since quite a while - I'm under the impression that you - still - don't take eq.(3.5) into account.
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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).
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I just looked up the Randall-Sundrum-paper and find myself confirmed. On p.5 in the first
paragraph (arxiv-version) or on p. 3372 (phys.rev.lett. 83, 17) they describe exactly the
procedure that I've suggested above. There is no division of actions whatsoever. So where is
the problem? Still no proof, I know, but it strongly suggest to try the same in the higher
dimensional case and for a more general class of metrics as in the G&M paper. In particular,
since the generalization of this procedure to higher dimension is straight forward if tedious. |
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Solkar
Anmeldedatum: 29.05.2009 Beiträge: 293
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Verfasst am: 08.07.2009, 11:37 Titel: |
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@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.
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 $M_D^{2}$).
Presumably this also would have been clear to you if you had cast away named "lazyness".
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Generally speaking - I'm not going to make up for you being "lazy" (your own words) by teXing your approach.
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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)
Best regards,
Solkar
P.S.: Btw "teXing" such analysis is generally considered a courtesy to the readers. |
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