Talk:Derivation of the Schwarzschild solution

You have that T_ab = 0 => R_ab = 0. But, after checking the link to the field equations, it looks like T_ab = 0 => G_ab = 0 => R_ab = (R/2) g_ab. What am I missing? 03:08, 27 Apr 2005 (UTC)


 * Have found the answer to my question. R_ab = (R/2) g_ab has only the solution R_ab = 0.  May I suggest that this less than obvious detail be added to the main article? Alfred Centauri 13:09, 27 Apr 2005 (UTC).


 * Requested details added to main article. --- Mpatel 11:12, 11 Jun 2005 (UTC)

Article virtually unintelligible to non expert reader
The article presupposes too much expert mathematical knowledge and needs to be rewritten to be understandable by a reasonably intelligent person with no specific mathematical knowledge. Either that or it needs to be taken off Wikipedia unless someone comes along who has the will to do this. All jargon and mathematical symbols, which abound in this article are unexplained and need to be qualified further so a non expert can follow what is being said. (eg. where theta = ...) Wikipedia is a general encyclpoedia and not a math textboom. This is unintelligible to a non mathematically trained individual. A rewrite would be in keepign with Wikipedia's policy of having intelligible articles. Thanks. — Preceding unsigned comment added by 176.26.237.167 (talk) 07:22, 15 September 2016 (UTC)


 * An encyclopedia is not supposed to have mathematicals derivation in the first place. We are not gonna go through the equivalent of the contents of 10 math courses just to do one article. 162.212.233.197 (talk) 00:57, 1 May 2024 (UTC)

Merger
This article mostly overlaps Schwarzschild solution. The actual derivation of the standard form of the solution can in fact be explained in a paragrapha, so we propose to merge with that article. ---CH 22:36, 3 March 2006 (UTC)


 * Hi CH. I agree that the derivation can be explained in a paragraph. In fact, I think there was a sketchy derivation in the main article about a year ago, but I realised that it would be better to have a separate article for a derivation. I created this article to specifically show readers how to actually derive an exact solution. I believe that this is very important and it conforms nicely with one of our objectives in the WikiProjectGTR, namely, to have articles representative of the work in GTR. Being such a standard procedure, I think it's worth showing an explicit derivation of such a historically important and still crucial solution in GR. I thought about doing the same for Kerr and Reissner-Nordstrom, but realised that this was unnecessary because this article gets the idea (of how to derive a metric) across with nice symmetry arguments. Thus, I would prefer that we not merge. MP  (talk) 08:28, 5 March 2006 (UTC)

I've put a thread heading at Talk:Schwarzschild solution to cover the merge, as that's where the "mergeto" template points discussion. If either of you wants to put a brief statement there, that would be great. --Christopher Thomas 20:42, 27 August 2006 (UTC)

i need explanation
in section Simplifying the components it is written that Choosing one of these hypersurfaces (the one with radius r0, say), the metric components restricted to this hypersurface (which we denote by \tilde{g}_{22} and \tilde{g}_{33}) should be independent of θ and φ (again, by spherical symmetry). but \tilde{g}_{33} depends on θ.


 * Hello there. I corrected my mistake. I meant to say 'unchanged under rotations of theta and phi' instead of independent of theta and phi' (see the assumption of spherical symmetry at the start of the article). Hope this clears up the confusion. MP  (talk) 22:50, 26 May 2006 (UTC)

Static
There is a mistake (or at least sloppy phrasing) in the article when it states that according to Birkhoff's theorem the solution is static. This is false - in fact the Schwarzschild solution is not even stationary inside the horizon. The correct statement of Birkhoff's theorem is that a spherically symmetric solution must be independent of the t-coordinate (and also asymptotically flat). Independence of t outside the horizon implies static but it doesn't mean this inside as t is spacelike there. JanPB 00:04, 31 January 2007 (UTC) (edited JanPB (talk) 04:28, 29 January 2011 (UTC))

Incorporate into Schwarzschild metric?
A query was made on Wikipedia talk:WikiProject Physics regarding the status of deriving the Schwarzschild solution. I suggest that the article be incorporated into the Schwarzschild metric but with a show/hide derivation box, such as in Hooke's law. MP (talk•contribs) 08:42, 26 October 2007 (UTC)

Coordinates in Christoffel symbols
In the section ‘Calculating the Christoffel symbols,’ another notation of the metric tensor is used than before. Besides the fact that in can be confusing if suddenly another notation is used, I have noticed that this notation uses another coordinate chart: $$\left(t, r, \theta, \phi \right)$$ instead of $$\left(r, \theta, \phi, t \right)$$ (which is given at the beginning of the article). Then, the Christoffel symbols are not obviously used to derive the vacuum field equations. They are contained within $$R_{ab}$$, but the article does not give how. I suggest the Christoffel symbols that can be derived from the notatin of the metric which has been used before should be employed and the derivation or at least derivation approach of the Ricci tensor included.

I have calculated the Christoffel symbols in coordinates $$\left(r, \theta, \phi, t \right)$$:
 * $$g_{ik} = \begin{bmatrix}

a & 0 & 0 & 0\\ 0 & r^2 & 0 & 0\\ 0 & 0 & r^2 \sin^2 \theta & 0\\ 0 & 0 & 0 & b \end{bmatrix}$$


 * $$\Gamma^0_{ik} = \begin{bmatrix}

a'/\left( 2a \right) & 0 & 0 & 0\\ 0 & -r/a & 0 & 0\\ 0 & 0 & -r \sin^2 \theta /a & 0\\ 0 & 0 & 0 & -b'/\left( 2a \right) \end{bmatrix}$$


 * $$\Gamma^1_{ik} = \begin{bmatrix}

0 & 1/r & 0 & 0\\ 1/r & 0 & 0 & 0\\ 0 & 0 & -\sin\theta\cos\theta & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$


 * $$\Gamma^2_{ik} = \begin{bmatrix}

0 & 0 & 1/r & 0\\ 0 & 0 & \cot\theta & 0\\ 1/r & \cot\theta & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$


 * $$\Gamma^3_{ik} = \begin{bmatrix}

0 & 0 & 0 & b'/\left( 2b \right)\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ b'/\left( 2b \right) & 0 & 0 & 0\end{bmatrix}$$

91.62.87.58 (talk) 10:59, 13 March 2012 (UTC)
 * I agree. All the notations: the metric, sign convention, parameterization - should be consistent in this article, otherwise it confuses the reader. It's not important to show all the different possibilities in the derivation. Setreset (talk) 05:24, 9 November 2012 (UTC)
 * I checked the symbols and made a correction. I am putting them instead of the current symbols (wp:bold...). Setreset (talk) 09:40, 10 November 2012 (UTC)
 * Agreed. When I started this article, I used the convention at the start. The person who wrote the helpful Christoffel symbols section used a different convention. I will stick to the original convention, check the Christoffel symbols and also flesh out some details of the Ricci tensor. Good suggestions. Thanks MPatel (talk•contribs) 22:29, 21 August 2013 (UTC)


 * There still seems to be some issues with the indexes. sometimes it is 0,1,2,3 and sometimes 1,2,3,4.  I'm far from an expert, so I'm not changing it.  Maybe this is traditional?  :-)  Dean p foster (talk) 18:44, 25 September 2013 (UTC)

A clear, useful article. However, it seems inelegant that the modern convention of associating t with the index 0 is not adhered to. It is rare that this convention is violated in current literature, for example in Physical Review D. Eddington used index 4 for t in the source for this article, however his book was written in 1923. 75.164.87.55 (talk) 18:14, 27 August 2014 (UTC)
 * I agree. Moreover, as mentioned by MPatel, the indexes are not consistent. I will change them in a week or so if no one objects and I remember. — Preceding unsigned comment added by Rondefalium (talk • contribs) 03:39, 8 March 2015 (UTC)

Explanation makeover
Section “Using the Weak-Field Approximation to find $$K$$ and $$S$$” doesn't look clearly explained to me. Questions that come to my mind reading it are, for instance, “Does the $$\approx$$ in $$g_{44}=K\left(1 +\frac{1}{Sr}\right) \approx -c^2+\frac{2Gm}{r} = -c^2 \left(1-\frac{2Gm}{c^2 r} \right)$$ mean the Schwarzschild solution is just an approximation?” or “How do you get from $$0=\delta\int\frac{ds}{dt}dt=\delta\int(KE+PE_g)dt$$ to $$g_{44}=K\left(1 +\frac{1}{Sr}\right) \approx -c^2+\frac{2Gm}{r} = -c^2 \left(1-\frac{2Gm}{c^2 r} \right)$$?”. If that wouldn't go too much into detail, it should be explained more thoroughly, I think, or at least provide some references to articles where similar situations are explained in detail. Unfortunately I'm not an expert on General Relativity or even physics, so I have to ask for these changes instead of changing the article on my own. --78.51.89.166 (talk) 21:21, 5 June 2015 (UTC)


 * No, the Schwarzschild solution is exact. Two formulas being approximately equal does not imply that they cannot be exactly equal.
 * The diagram in that section gives a better idea than the text of how one derives the formula. JRSpriggs (talk) 10:01, 7 June 2015 (UTC)

Reasoning incorrect
In Section "Alternate derivation using known physics in special cases," it states that $$\ddot{r} = -MG/r^2$$ is the acceleration of gravity. However, the dots indicate differentiation with respect to proper time, not coordinate time. Using the chain rule, $$\ddot{r} = \frac{dt}{d\tau}\frac{d}{dt}\left(\frac{dr}{dt}\frac{dt}{d\tau}\right) = \frac{d^2r}{dt^2}\dot{t}^2 + \frac{dr}{dt}\frac{d}{dt}\left(\frac{dt}{d\tau}\right)\dot{t},$$ or, in the special case when $$dr/dt = 0$$, this simplifies to $$\ddot{r} = \frac{d^2r}{dt^2}\dot{t}^2.$$ Plugging this into the indicated E-L equation, you find that the $$\dot{t}^2$$ cancels, and you get $$A(r) = 1$$, which lets you recover Newtonian gravity (unsurprisingly, since Newtonian gravity is all that was assumed in this part of the derivation), but not the full Schwarzschild metric. — Preceding unsigned comment added by Smcsweeney (talk • contribs) 05:24, 16 March 2022 (UTC)

Problem with the invariance of the metric
Section "Diagonalising the metric". When you write the transformation law for $$g_{\mu 4}$$, it should be understood: $$ g'_{\mu 4} (x') = \frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^4}g_{\alpha\beta}(x)= -g_{\mu 4}(x) $$, while at the same time the invariance tells you that $$ g'_{\mu 4}(x') = g_{\mu 4}(x') $$. This leads to the conclusion that $$g_{\mu 4}(x') = -g_{\mu 4}(x)$$, but I fail to see how to go further in the reasoning without any aditionnal assumption.