Talk:Einstein–Hilbert action

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    The Einstein-Hilbert action as stated in the article is S[g]=k\int d^nx \sqrt{|\det(g)|} R. Unfortunately, there is no hint on the domain over which the integral is taken nor if g is varrying with x. Am I missing something or is the article missing something? 84.160.232.20 21:03, 10 September 2005 (UTC)

    The article is missing something, I guess. The integral is taken over (a neighborhood of) spacetime and g does vary. The volume element is just d^4 x \, \sqrt{-g} for a four dimensional Lorentzian manifold, so this just says we integrate the Ricci scalar over spacetime. And in fact there is something to say about boundary terms.---CH (talk) 18:57, 21 October 2005 (UTC)

    The integral is taken over the whole manifold, i.e. the whole space-time. Of course, the determinant of the metric g is varying with x (otherwise it should have been written in front of the integral) as it describes the shape of space-time, the same holds for the scalar curvature R. So for a flat Minkowski space-time, that's the shape of space-time without or very weak gravity, g does not depend on x, but a curved space-time has varying g. The usual approach to boundary terms in undergraduate texts is to simply not discuss them. The usual argument given, the article says the same, BTW, is that the variation is assumed to vanish at the boundary. Then the contribution of the boundary terms is multiplied by a zero. There is indeed more to be said about boundary terms, see for example AdS/CFT correspondence, but this goes far beyond a Wikipedia article, I think. The only thing I consider unfortunate is that the conventions seem to differ between different articles, for example the definition of the stress-energy tensor given there and in Einstein-Hilbert action.

    Contents

    [edit] Source for claim Einstein consulted Hilbert?

    What is this claim that Einstein went to Hilbert to get the field eqiuations? Hilbert invited Einstein to Gottingen to give six 2-hour lectures in June-July 1915 on general relativity. This was prompted by Hilbert's reading Einstein/Grossmann "Outline" papers in 1914 where the idea of linking space-time curvature to gravity was introduced (Sauer 1999, Arch. Hist. Exact Sci, v53, p529-575). Hilbert was always seeking physicists for problems in theoreticfal physics (Hilbert had previously invivted Einstein to talk on the kinetic theory, which Einstein declined) - Einstein did not go to Hilbert for help. Sauer is a well–researched and foot-noted paper. E4mmacro 05:10, 10 March 2006 (UTC)

    Didn't Hilbert derive the field actions from the action principle first? Einstein "derivation" of the field equations was more of a guess based on all the previous versions he had tried, not a derivation in any mathematical sense. When was it first called the Einstein-Hilbert action? E4mmacro 06:59, 26 March 2006 (UTC)

    Way too much ink has been spilled on this, but IMO standard biographies, recent history of science papers (by reputable researchers), and available primary sources make clear why the standard name is indeed appropriate. In any case, as far as WP is concerned, I think it suffices to agree that this is the standard name; this is not a forum for debating whether the standard name is somehow unfair to anyone. ---CH 01:22, 27 March 2006 (UTC)

    Wasn't trying to change it or debate it. Just asking for confirmation that this is the standard name (and out of curiosity, when that standard name was first used). E4mmacro 09:42, 27 March 2006 (UTC)

    [edit] Definition of ln(gm,n) in "Variation of the determinant"

    Suppose we take the flat minkowski metric g = diag(1,-1,-1,-1) How is ln(g) defined? as g is diagonal ln(g) = diag(ln(1),ln(-1),ln(-1),ln(-1)) but what is ln(-1) ? —The preceding unsigned comment was added by 141.20.50.155 (talk) 15:07, 4 January 2007 (UTC).

    I removed the reference to the logarithm. JRSpriggs 05:06, 5 January 2007 (UTC)

    You can do it in the following way:

    \delta g = \frac{\delta g}{\delta g_{ab}} \delta g_{ab} = \Delta_{ab} \delta g_{ab} = g \frac{\Delta_{ab}}{g} \delta g_{ab} = g g^{ab} \delta g_{ab}

    where:

    Δab is the adjugate matrix of g.

    The method with log is correct. We may define log on the whole complex plane, e.g.

    \log \left( -1 \right) = i \pi because \exp \left(i \pi \right) = -1

    ... —Preceding unsigned comment added by Eryk.schiller (talk • contribs) 23:18, 6 May 2007

    [edit] Variation of curvature calculation is wrong

    I believe this statement is very wrong.

    Notice that \delta\Gamma^\rho_{\nu\sigma} is a tensor. So we can transform to a coordinate system where \Gamma^\rho_{\nu\sigma} is zero. This causes the terms without partial derivatives to vanish. Also, in the terms with partial derivatives, they become covariant derivatives.

    The first problem is that \Gamma^\rho_{\nu\sigma} is not a tensor so it's not clear to me why its variation would be a tensor. Second, by transforming \Gamma^\rho_{\nu\sigma} to zero, you get the trivial result that curvature, {R^\rho}_{\sigma\mu\nu} is zero. Ie, you are flattening the space. Finally, one doesn't need to perform this dubious transformation to show \delta R^r{}_{mln} = \nabla_l (\delta \Gamma^r_{nm}) - \nabla_n (\delta \Gamma^r_{lm}) . The variation of the definition of the curvature tensor, a few lines above (plus some relatively mild crunching) yields the desired result with no need for these sorts of assumptions. -- KarlHallowell 22:09, 24 April 2007 (UTC)

    From Christoffel symbol#Change of variable, we get the transformation law of a connection is:
    \overline{\Gamma^k {}_{ij}} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r {}_{pq}\, \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial y^i \partial y^j} \
    This holds for both the pre-variation connection and the post-variation connection. If you subtract them to get the transformation law for the variation itself, the second terms on the right hand side cancel out. So the variation of the connection is indeed a tensor, even though the connection itself is not a tensor.
    As far as the transformation to a coordinate system where the connection is zero, I am only transforming it to zero at one event. Thus the derivatives of the connection do not have to become zero. Thus the curvature can be non-zero. So you are mistaken on both counts. The calculation comes out of a reference work (see page 500 of Gravitation (book)), I did not make it up. It is a standard way of simplifying the work. JRSpriggs 11:39, 25 April 2007 (UTC)
    I should mention that the reason you can choose a coordinate system where the connection at one event is zero is that the second term on the right side of the transformation law can be made to cancel the first term at that event by a suitable choice of the functions ym. JRSpriggs 03:42, 1 May 2007 (UTC)


    Just happened by. I retract my objections to the above calculations. -- KarlHallowell 22:52, 30 October 2007 (UTC)

    [edit] Spacetime vs. Space-time

    I just undid part of a edit by JRSpriggs replacing all occurrences of spacetime with space-time. Both forms appear. (Both have about 6 million hits on google for whatever thats worth.) The consensus on wikipedia seems to be to go with the first form. At least the spacetime article is listed under that name. So I think we should just keep it at that form.

    I also removed a remark about which formalism we would be assuming which was out of place in the lead section. (If it is deemed necessary it can be reinstated at the start of the derivation section. (TimothyRias (talk) 15:04, 31 March 2008 (UTC))

    Mentioning "The Palatini formulation of general relativity" in the lead creates the false impression that it is going to be used in the remainder of the article. That is why I tried to counteract that impression by adding another sentence. If you want to simply not mention the Palatini formulation at all, that would be acceptable. By the way, my spelling checker does not like the unhyphenated form "spacetime". JRSpriggs (talk) 02:57, 1 April 2008 (UTC)

    [edit] Constant coefficient k

    Currently, the article uses k = \frac{c^4}{16 \pi G}. This is contrary to all the literature with which I am familiar. We should replace the k in this article with 1/k everywhere in the article so that we would have k = \frac{16 \pi G}{c^4}. JRSpriggs (talk) 07:40, 2 April 2008 (UTC)

    Fine by me. (TimothyRias (talk) 07:53, 2 April 2008 (UTC))
    I now see that I need to think about this some more. k = \frac{8 \pi G}{c^4} is the constant which usually appears on the right side of the Einstein field equations. Notice the "8" instead of "16". And I am not sure about the correctness of constant factor in
    T_{\mu\nu}:= -2 \frac{1}{\sqrt{-g}}\frac{\delta (\sqrt{-g} \mathcal{L}_\mathrm{M})}{\delta g^{\mu\nu}}.
    So I will have to think some more. JRSpriggs (talk) 12:42, 2 April 2008 (UTC)
    The constants are the ones used by Carroll, as is the convention used for the constant factor in the definition of the stress-energy. It is important though to have the constant precede the Einstein-Hilbert term (whether this is k or 1/k is a matter of taste.) Firstly, because is gives the action the right units (which should be [Energy][Time]). Secondly, placing it before the matter term is an invitation to mistakes if more than one matter term would be present. (TimothyRias (talk) 13:53, 2 April 2008 (UTC))
    Yes, the units must be such that the action has units of joule·seconds. The action of an isolated massive particle should reduce to \int { ( -m c^2 + \frac {m v^2}{2} ) d t } in the non-relativistic limit. JRSpriggs (talk) 10:58, 3 April 2008 (UTC)