Talk:Einstein–Hilbert action

Sign of the Action Again

The sign of the action is shown as negative in the introduction but positive in the section "Derivation of Einstein's Field Equations". One or the other should be changed. There seems to be an edit war here revolving around signature (+ - - -) vs (- + + +). I really don't care which is used, but can this article at least be consistent? I'd change it myself, but, being incompetent in this area, I really don't want possibly to mask a greater, more subtle problem by correcting its most obvious manifestation.

201.215.63.217 (talk) 20:50, 12 September 2010 (UTC)Thinkor

Mathematicians usually use (-+++) for good reasons: then the set {x\in R^{3,1}|g(x,x)=-1} is a model for hyperbolic space, i.e. the riemannian 3-dimensional manifold of constant sectional curvature -1.

Physicists usually use (+---) for other good reasons. This is the standard convention in physics.

Which sign you choose, is in fact irrelevant, but it should be consistent. In any case det g is negative, thus |det g|= - det g.

79.209.86.154 (talk) 19:16, 28 October 2010 (UTC)

Domain of integration

The Einstein-Hilbert action as stated in the article is $S[g]=k\int d^{n}x{\sqrt {|\det(g)|}}R$. 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.

79.209.86.154 (talk) 19:07, 28 October 2010 (UTC)

The contribution of the person above is partially correct. That the integral is taken over the whole manifold is certainly a good claim. The trouble is that it is not clear whether the integral will actually converge when integrated over the whole spacetime. This is not just a theoretical small technical issue as there are many examples where the integral diverges, but which are physically relevant, e.g. divergence will occur in almost all stationary spacetimes. The solution to the dilemma is the following. One is not really interested in the value of the action. The only thing we need are variation formulas. And these variation formulas will be the Einstein equations. We should ask at first. Variation with respect to what? There are several natural classes of variation to look at, e.g. (1) compactly supported variations of the metric (2) rapidly (to be specified) decaying variation (3) arbitrary variations To vary with respect to (3) is bad. In order to derive the Einstein equation as an Euler-Lagrange equation to the action, one has to partially integrate, and this is not possible if one admits arbitrary variations. Admitting rapidly decaying variations is also sometimes important, e.g. if one wants to derive the Arnowitt-Deser-Misner mass of an asymptotically flat spacetime. But in this context, this is not the right choice either. We have to work with (1), i.e. compactly supported function. Thus our first goal is to make sense out of (d/ds) S(g+th) for a compactly supported symmetric 2-tensor h. Assume that the integral in the definition of S(g) converges. Then the expression (d/ds) S(g+th) is independent on the domain of intergation, provided that the domian of integration contains the support of h. It is thus consistent to define: g is a stationary point of the functional $g\mapsto S(g)$ if for any compactly supported h and any relatively compact open domain of integration \Omega containing the support of h, one has (d/ds) S(g+th)=0 for s=0. As explained above it is sufficient to check this for one single domain for each compactly supported h. Now one can prove using partial integration that g is a stationary point in this sense if and only if it satisfies the Einstein equation which thus can be regarded as the Euler-Lagrange equation of this functional.

To summarize: the important fact here is that we made sense to the variational formulas of the functional although it is not clear that the functinal itself actually is well-defined, even worse the integral will diverge in most physically relevant situations. But this divergence does not matter at all, as we only consider compact variations anyhow. I emphasize: what we did above is mathematically absolutely rigorous, but clearly: what I explained was too short to be mathematically complete, it is just a sketch of the idea.

To say that we consider variation vanishing at the boundary is certainly a good choice. The boundary terms can not be neglected and give very unpleasant contributions.

Literatur: For the variation of the functional see e.g. Besse's book on Einstein manifolds, Springer-Verlag. How rapidly decaying variations leed to a definition of the ADM-mass is e.g. well explained in the article of Parker and Lee on "The Yamabe problem" which was published in the Bulletin of the American Math. Society, but there are certainly other, more original sources for this. —Preceding unsigned comment added by 79.209.86.154 (talk) 19:06, 28 October 2010 (UTC) 79.209.86.154 (talk) 19:09, 28 October 2010 (UTC)

Is the integral calculated over the local coordinate chart in some neighbourhood, hence the need for the volume element? Troubling that would mean that the boundary conditions can only be defined up to the edge of the chart; and there is also the thorny question of ascertaining the existence of a manifold given an atlas, and smoothly related metrics on the charts of the atlas. Is compatibility of the metric with the atlas sufficient or are there additional necessary conditions and tests of consistency? 68.146.90.105 (talk) 04:17, 30 January 2016 (UTC)

References

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)

Definition of $ln(g_{m,n})$ 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:

$\delta g={\frac {\delta g}{\delta g_{ab}}}\delta g_{ab}=\Delta _{ab}\delta g_{ab}=g{\frac {\Delta _{ab}}{g}}\delta g_{ab}=gg^{ab}\delta g_{ab}$

where:

$\Delta _{ab}$ 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

... the perforated complex plane - ln(0) is an essential singularity. — Preceding unsigned comment added by 131.111.184.95 (talk) 11:25, 26 March 2013 (UTC)

Variation of curvature calculation is wrong

I believe this statement is very wrong.

Notice that $\delta \Gamma _{\nu \sigma }^{\rho }$ is a tensor. So we can transform to a coordinate system where $\Gamma _{\nu \sigma }^{\rho }$ 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 $\Gamma _{\nu \sigma }^{\rho }$ is not a tensor so it's not clear to me why its variation would be a tensor. Second, by transforming $\Gamma _{\nu \sigma }^{\rho }$ 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 $\delta R^{r}{}_{mln}=\nabla _{l}(\delta \Gamma _{nm}^{r})-\nabla _{n}(\delta \Gamma _{lm}^{r})$. 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 y^m. 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)

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)

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 $\int {(-mc^{2}+{\frac {mv^{2}}{2}})dt}$ in the non-relativistic limit. JRSpriggs (talk) 10:58, 3 April 2008 (UTC)

OK, I have belatedly replaced $k\rightarrow {1 \over 2\kappa }\,$ so that $\kappa ={8\pi G \over c^{4}}\,.$ I also put a factor of 1/c^2 into the definition of g. This has no effect on any of the calculations. It merely ensures that g=-1 when in Cartesian coordinates and SI units in special relativity, because ${\sqrt {-g}}=1\,$ must hold in special relativity for backward compatibility. JRSpriggs (talk) 08:23, 8 August 2008 (UTC)

Should it not be mentioned that you included a factor of c in the determinant of g? I find this notation very confusing.

Bakken (talk · contribs) undid the units correction which I described in my previous comment here. His version follows the sources I have read, but it makes no sense unless you choose to ignore factors of c entirely. JRSpriggs (talk) 06:22, 30 April 2014 (UTC)

The sign of the action

I have "Feynman lectures on gravitation" in front of me right now and on page 136 (equation (10.1.2)), the action for the gravitational field is given as

$S_{g}=-{\frac {1}{2\lambda ^{2}}}\int d^{4}xR{\sqrt {-g}}\;.$

It seems it has the opposite sign compared to the action in this article. My correction of the sign in the article to make it consistent with Feynman had been deleted as "unhelpful" by JRSpriggs. Could the authors of this article please explain me what is wrong with Feynman? :) Bakken (talk) 20:47, 12 November 2009 (UTC)

I think I have found where the mistake is. The article postulates (in "Derivation of Einstein's field equations") that

${\frac {1}{2}}{\sqrt {-g}}T_{ab}\delta g^{ab}=-\delta ({\sqrt {-g}}L)\;,$

from which it follows that the variation of the action for the matter is

$\delta S=+{\frac {1}{2}}\int T^{ab}\delta g_{ab}{\sqrt {-g}}d^{4}x.\;\;(1)$

The seems to be wrong. Indeed, let us consider a particle with mass $m$. Its action is given as

$S=-m\int ds=-m\int {\sqrt {g_{ab}dx^{a}dx^{b}}}.$

The minus sign here can be easily checked in the non-relativistic limit. [Indeed,

$S=-m\int dt{\sqrt {1-v^{2}}}\to \int dt({mv^{2} \over 2}-m),$

which is the correct non-relativistic action. Bakken (talk) 18:41, 7 February 2010 (UTC)] Now, a variation of this action with respect to the metric tensor apparently gives

$\delta S=-{\frac {1}{2}}\int mu^{a}u^{b}\delta g_{ab}ds\;.$

According to (1) the metric tensor for a particle with the mass $m$ is then given as

$T^{ab}=-mu^{a}u^{b}.\;\;(2)$

This is obviously wrong. Indeed, if the particle is at rest its energy, the $T^{00}$ component, is $+m$ not $-m$, as it follows from (2).

So, it seems like Feynman was right after all... ;) Bakken (talk) 21:49, 12 November 2009 (UTC)

Well, nobody cares to answer? I'm starting changing the text then -- please, do not start wikipedia:edit warring. Bakken (talk) 13:09, 14 November 2009 (UTC)

Your expression for the action of a particle is incorrect. See Lagrangian#General relativistic test particle for the correct expression. The minus sign under the radical accounts for your sign error. It is required because we use the (-+++) signature following Gravitation (book). JRSpriggs (talk) 14:31, 7 February 2010 (UTC)

P.S. Richard Feynman was a genius but he was very careless about keeping track of constant factors. Also remember many articles will have equations which differ from what you find in your text books because their authors have selected conventions from different sources which unfortunately use different definitions than your texts. JRSpriggs (talk) 14:46, 7 February 2010 (UTC)

Are you saying that the equation (10.1.2) in Feynman lectures on gravitation is wrong because of his "very careless" approach to writing books? Man, that is quite an observation you've made. Could you, perhaps, substantiate your claim? I am sure the editor will be *very interested* :) Bakken (talk) 18:33, 7 February 2010 (UTC)

Your expression for the action of a particle is incorrect. :)

well, it is equation (8,1) in Landau & Lifshitz, Classical theory of fields. Incorrect, ah? Bakken (talk) 18:33, 7 February 2010 (UTC)

It is required because we use the (-+++) signature. Well, *we* use (+---) signature after Weinberg and Landau. Since the Sign convention is apparently a matter of grave importance, the only solution, acceptable for both particle physicists and relativity physicists, would be to write every expression twice with each signature. So be it. Bakken (talk) 18:33, 7 February 2010 (UTC)

I think k should have a c^3, not c^4 in it. See ee.g. Landau and Lifshitz, Classical Theory of Fields, Art 93, p268 (4th Rev English Ed) - Dasbabu —Preceding unsigned comment added by 96.51.51.120 (talk) 00:34, 22 May 2011 (UTC)

More on the sign of the action

I'm not saying the change made by Bakken to the sign of the action is wrong, but now we have different signs for the "action" and the "full action". That doesn't seem right. Thinkor (talk) 07:37, 7 February 2010 (UTC)

I didn't realise it at first, but it seems that JRSpriggs uses the space-like metric signature which is opposite to what is used e.g. in wikipedia articles Schwarzschild_metric, Reissner–Nordström_metric and Kerr_metric. It seems that in this opposite signature convention the sign of the action might be opposite to the time-like convention. Indeed, the energy-momentum tensor must have the same sign in both conventions since the energy (difference) is an observable quantity. Now, the energy-momentum tensor is proportional to the variational derivative of the action $T_{ab}\propto \delta S/\delta g_{ab}$. Therefore, when the sign convention is changed, $g_{ab}\to -g_{ab}$, the action has to change the sign as well in order to preserve the sign of the energy-momentum tensor. Bakken (talk) 20:13, 7 February 2010 (UTC)

How about defining a constant for the spacetime signature up front, such as $\eta =1$ for {1,-1,-1,-1} and $\eta =-1$ for {-1,1,1,1}, and work it through from there, so it's signature agnostic? --Scientryst (talk) 09:16, 1 March 2010 (UTC)

Please do this. It would really help to avoid much confusion. —Preceding unsigned comment added by 138.251.106.33 (talk) 16:50, 20 October 2010 (UTC)

Confusing pre-factor doesn't matter without ${\mathcal {L}}_{\mathrm {M} }$

Specifically the ${1 \over 2\kappa }$ factor doesn't matter without taking the ${\mathcal {L}}_{\mathrm {M} }$ into account. Confusing people about what it is for. On the plus side, it does make people who notice it look better at it. 82.169.255.79 (talk) 19:47, 25 November 2010 (UTC)

Whence the action?

The article doesn't do much to explain where the action comes from. Can someone explain in a reasonably simple but not vague way where the action comes from. The reference to the article on the principle of least action did little to clear things up for me. Why is it curvature as measured by the Ricci scalar that is being minimized? What are the mysterious matter fields? What is the point of the 2k? If you minimize without the 2k don't you get the same result? Isn't it just a constant? I am assuming here that there is a Lagrangian multiplier involved in the term for the matter fields, but since there is no description of the matter fields I have no idea. What is the sqrt(-g) for?

The article doesn't really seem to be about the action, but about the derivation of the field equations from the action. I believe the action lies at the heart of understanding GR, but I can't find an explanation for it. 200.83.84.63 (talk) 03:09, 16 August 2013 (UTC)

See Lagrangian#Newtonian gravity for the Lagrangian density used to get the action which leads to the Newtonian gravitational field. Also see Gauss's law for gravity. Since Newton's theory of gravity is very very close to the truth as confirmed by extensive observation, the actual Lagrangian density should reduce to the Newtonian approximation in the non-relativistic limit. The Newtonian gravitational potential should be replaced by the metric tensor (general relativity) to fulfill the equivalence principle which is known empirically to be characteristic of gravity. And, as Einstein realized, any reasonable theory must be invariant under general coordinate transformations (called "general covariance"). The Einstein–Hilbert action is by far the simplest action which satisfies these criteria.

I am sorry that there is no easy way to understand this. But the important point is that this action leads to Einstein's theory of gravity (general relativity) which has been confirmed by experiments. However, other theories are still possible — see alternatives to general relativity.

See Ricci tensor for an explanation of its relationship to curvature. What the matter fields are is outside the scope of this theory and article — for our purposes, they are just a black box. The constant is needed to get the units right when combining with matter fields or comparing to Newtonian gravity. The value of the constant is derived at Einstein equations#The correspondence principle. There is no Lagrange multiplier. The square root is needed to make the integrand a scalar density which it must be to produce a generally covariant theory. JRSpriggs (talk) 04:26, 16 August 2013 (UTC)

To further clarify, the constant 1/(2κ) is a measure of the stiffness (resistance to bending) of the spacetime continuum. Sqrt{-g} is the amount of physical hypervolume in a unit volume of coordinate space. JRSpriggs (talk) 02:43, 17 August 2013 (UTC)

The (1/2κ) is needed to make the action integral S = ∫ (1/2κ) R √|g| dⁿx dimensionally correct since, otherwise, the integral would have dimension [∫ R √|g| dⁿx] = √(Aⁿ⁻²), for n-dimensional spacetime, where A = [g_{μν} dx^μ dx^ν] is the dimension of the line element, which is normally taken as A = L² (for L = length). In 4D, n = 4, that works out to [∫ R √|g| d⁴x] = A = L². Since you already have, for the action, [S] = ML²/T (for M = mass, T = time), then this requires a coefficient κ with [κ] = AT/ML² (= T/M in 4D). The particular form - as 1/2κ, instead of (say) k - is conventional, written that way purely a matter of convenience. Oh, and I noted elsewhere, the dimensions for κ mean that the expression given in the article is wrong. It can't be κ = 8πG/c⁴, since [8πG/c⁴] = T²/ML; but instead has to be κ = 8πG/c³, since [8πG/c³] = T/M. The expression in the article only makes sense if the line element is given dimensions A = LT, which I don't think is ever done. — Preceding unsigned comment added by 2603:6000:AA40:1EF2:222:69FF:FE4C:408B (talk) 19:19, 12 August 2020 (UTC)

SI units

action has units Js, R has 1/m² and Kappa k has 1/N, so I suppose, the integral is supposed to contribute m³s. x°°=t thus is here in units s, which is unusual and confusing, I would prefer x°°=ct in units m thus adding a divisor c which is multiplying c k. Ra-raisch (talk) 19:13, 21 December 2016 (UTC)

The real problem here is that units of ${\sqrt {-g}}$ should be 1 for compatibility with special relativistic formulas, but it actually is c (length over time). I tried to fix that some time ago (by effectively replacing ${\sqrt {-g}}$ by ${\sqrt {\frac {g}{\eta }}}={\frac {\sqrt {-g}}{c}}$ using the determinant of the Minkowski metric), but I was reverted because the formula I put in was not the same as the "authoritative" sources. JRSpriggs (talk) 01:36, 23 December 2016 (UTC)

hmm, I think, eta is of units 1 just like g. It is just a special (uncurved) case of g whereas g stands general for metric which may be curved. But the units are the same: none which is 1 (value -1). Ra-raisch (talk) 09:40, 25 December 2016 (UTC)

I see, there are two or even three conventions |eta|=eta.°°=-c or |eta|=eta.°°=-1 or |eta|=-eta.°°=-1, I prefer matrices with uniform units. Ra-raisch (talk) 18:22, 3 January 2017 (UTC)

In general, the units for the metric are set by the units for the line element. I did the analysis in a couple comments above, so I won't repeat it here. The key points to emphasize are: (1) the metric determinant is generally not dimensionless, and (2) its dimensions depend on how many dimensions spacetime has. The relevant quantity you should be asking the dimensions of, actually, is the volume n-form √|g| dⁿx. Its dimensions in n-dimensional space time are √(Aⁿ), where A is the dimension of the line element. With the usual setting A = L², that works out to Lⁿ, or in 4D space-time, just L⁴. If you use ct for the time-like coordinate, instead of t, then you can get away with having g be dimensionless. But for the natural quantities (t for time, (x,y,z) for spatial coordinates) g will have dimensions (L/T)² and |g| = c² in the flat-spacetime case of the Minkowski metric. — Preceding unsigned comment added by 2603:6000:AA40:1EF2:222:69FF:FE4C:408B (talk) 19:31, 12 August 2020 (UTC)

Physical context of vanishing surface term

The Universe seems to be finite (standard ΛCDM model). After this, how we are able (conceptually) to take the limits of the integral of total derivative term to be infinite? — Preceding unsigned comment added by 94.66.57.122 (talk) 10:21, 22 October 2017 (UTC)

I do not understand your question. Could you please be more specific. JRSpriggs (talk) 01:15, 23 October 2017 (UTC)

I think it is a valid question, I suppose you refer to the derivation of equation of motion. That explanation of the boundary term should be made precise mentioning the case of a spacetime with boundaries, and there should be a link to the Brown-York stress tensor page. --PhysicsLine (talk) 23:01, 11 March 2018 (UTC)

Your question is still unclear. What is the "Brown-York stress tensor"?

However, you seem to be asking about the boundary term vanishing "at infinity". So I reworded that sentence. I hope that helps. JRSpriggs (talk) 02:48, 12 March 2018 (UTC)

The power of c in the coupling coefficient is not correct and needs to be fixed.

Actually, it would be more appropriate to say that there is no universal consensus. Einstein uses c² in The Meaning of Relativity, the Wikipedia articles use c⁴. I've also seen c³ (e.g. "On the Vielbein Formulation of General Relativity, GRC (2008) 40: 1913-1945). The most natural selection is c³, not c⁴, as I'll explain here, but it depends on what convention is used for the Minkowski metric.

Using L, T, M respectively for length, time and mass, and [X] to denote the dimensions of X; then [action] = ML²/T. The Einstein-Hilbert action is proportional to ∫ R vol(g), where vol(g) ≡ √|g| d⁴x is the invariant 4-form derived from the metric and R the curvature scalar. The multiplying coefficient is proportional to the inverse of the coupling coefficient, κ.

Setting, for the moment, A = [g_{μν} dx^μ dx^ν] for the line element, then [R] = 1/A and [vol(g)] = A². Thus, [∫ R vol(g)] = A, so one must therefore have [κ] = AT/ML². In addition, we have, for Newton's gravitational coefficient [G] = L³/MT² and light speed [c] = L/T. Finally, we have for the Minkowski metric coefficient [η₀₀] = A/T². Thus, [η₀₀ G/c⁵] = AT/ML², which matches what's required. So, assuming |η₀₀| is a power of c, the value of the coupling coefficient is κ = ±η₀₀ 8πG/c⁵; the matter of the sign is discussed elsewhere, so I won't touch on that here.

The only way to get a coupling coefficient with c⁴ is to adopt A = LT, which is normally not done. Usually, one adopts either A = L² to make the metric a distance metric, or sometimes A = T² to make it a proper time metric. In the respective cases, the power of c attached to the coupling coefficient is c³ or c⁵.

In the field equation G_μν + Λ g_μν = κ T_μν, for the cosmological coefficient, one must adopt [Λ] = 1/A, for this to match. Normally, the coefficient is taken as an inverse length squared, which is consistent with A = L² and η₀₀ = ±c². Thus, the most natural selection for the coupling coefficient is actually κ = ±8πG/c³, not κ = ±8πG/c⁴. — Preceding unsigned comment added by 2603:6000:AA40:1EF2:222:69FF:FE4C:408B (talk) 00:41, 12 August 2020 (UTC)

What is 1 integral with 4 differentials?

There must be 1 integral with 1 differential or 4 integrals with 4 differentials. The limit interval must be written. What is 1 integral with 4 differentials? Wikipedia does not explaine this. If non-standard designations are used, then they must be described. Voproshatel (talk) 07:02, 19 August 2022 (UTC)

It is an integral over a four-dimensional space, so it could be regarded as a quadruple integral with variables (x, y, z, t). But writing it that way would just be wasting the reader's time with unnecessary and obvious stuff. JRSpriggs (talk) 07:53, 19 August 2022 (UTC)