Talk:Einstein–Hilbert action

message: 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)
thread_title: Sign of the Action Again

message: 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)
replies: text: 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)
thread_title: Domain of integration

message: 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)
replies: text: 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)
thread_title:  Source for claim Einstein consulted Hilbert? 

message: 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)
replies: text: 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)
thread_title:  Spacetime vs. Space-time 

message: 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)
replies: text: 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)
thread_title:  More on the sign of the action 

message: 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)
thread_title:  Confusing pre-factor doesn't matter without <math>\mathcal{L}_\mathrm{M}</math> 

message: 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)
replies: text: 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)
replies: text: 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)

text: 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)

text:  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)
thread_title:  SI units 

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

message: 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)
replies: text: 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)
thread_title:  What is 1 integral with 4 differentials?