Talk:Metric tensor (general relativity)

This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Physics Mid‑importance This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles Mid This article has been rated as Mid-importance on the project's importance scale.

I cut out the material on the volume form which properly belongs at volume form. Here it is for reference:

Let [g] be the matrix of elements ${\displaystyle g_{\mu \nu }}$. Matrix [g] is symmetric, so due to a corollary of the spectral theorem, there exists an orthogonal transformation matrix Λ which diagonalizes [g], e.g.

${\displaystyle D=\Lambda ^{\top }[g]\Lambda }$

where D is a diagonal matrix whose diagonal elements are eigenvalues of [g]: ${\displaystyle D_{\alpha \alpha }=\lambda _{\alpha }}$. (Note that Λ can be chosen so that the eigenvalues are in numerical order, D₀₀ being the smallest.) Then there is a diagonal matrix V which "unitizes" D, i.e. which applies the mapping ${\displaystyle \lambda _{\alpha }\mapsto {\mbox{sgn}}(\lambda _{\alpha })}$ to the diagonal elements of D. Such matrix V has diagonal elements

${\displaystyle V_{\alpha \alpha }=\left\{{\begin{matrix}{1 \over {\sqrt {|\lambda _{\alpha }|}}}&\quad {\mbox{if}}\,\lambda _{\alpha }\neq 0\\0&\quad {\mbox{if}}\,\lambda _{\alpha }=0\end{matrix}}\right.}$

Then

${\displaystyle [\eta ]=V^{\top }\Lambda ^{\top }[g]\Lambda V}$

and for a given manifold, the trace of [η] will be the same for all points and is referred to as the signature of the metric. (A signature of +2 is synonymous with a signature of (− + + +). ) This matrix [η] has the components of the Minkowski metric, which means that the manifold is, at each one of its points, locally smooth.

The matrix ${\displaystyle (V\Lambda )^{\top }}$ is a Jacobian (a multivariate differential, or push forward) which transforms [η] to [g],

${\displaystyle [g]=V\Lambda [\eta ]\Lambda ^{\top }V^{\top }}$

and taking determinants

${\displaystyle g:={\mbox{det}}([g])={\mbox{det}}\,(V\Lambda )\,{\mbox{det}}([\eta ])\,{\mbox{det}}(\Lambda ^{\top }V^{\top })}$

${\displaystyle ={\mbox{det}}^{2}(V\Lambda )\,{\mbox{det}}([\eta ]),\ }$

${\displaystyle g=-{\mbox{det}}^{2}(V\Lambda ),\ }$

${\displaystyle {\mbox{det}}(V\Lambda )={\sqrt {-g}},}$

but due to a property of diffeomorphisms, a volume element ${\displaystyle dx^{0}dx^{1}dx^{2}dx^{3}}$ whose factors are components of an orthonormal basis (locally), when transformed to components ${\displaystyle dx^{\bar {\mu }}}$, has the determinant of the Jacobian matrix J as conversion factor:

${\displaystyle G=dx^{0}dx^{1}dx^{2}dx^{3}={\mbox{det}}(J)\,dx^{\bar {0}}dx^{\bar {1}}dx^{\bar {2}}dx^{\bar {3}}.}$

See also volume form.

-- Fropuff 18:02, 22 February 2006 (UTC)[reply]

Just as well. There seem to be several mathematical errors in the matrix calculations. JRSpriggs (talk) 00:21, 5 April 2016 (UTC)[reply]

See Sign convention ---CH 01:54, 25 May 2006 (UTC)[reply]

The anecdote was deemed too long, so it was deleted. However, it is in the edit history. --50.39.98.212 (talk) 23:42, 1 April 2022 (UTC)[reply]

${\displaystyle g_{{\bar {\mu }}{\bar {\nu }}}={\frac {\partial x^{\rho }}{\partial x^{\bar {\mu }}}}{\frac {\partial x^{\sigma }}{\partial x^{\bar {\nu }}}}g_{\rho \sigma }=\Lambda ^{\rho }{}_{\bar {\mu }}\,\Lambda ^{\sigma }{}_{\bar {\nu }}\,g_{\rho \sigma }.}$

should perhaps be:

${\displaystyle g_{{\bar {\mu }}{\bar {\nu }}}={\frac {\partial x^{\rho }}{\partial x^{\bar {\mu }}}}{\frac {\partial x^{\sigma }}{\partial x^{\bar {\nu }}}}g_{\rho \sigma }{\overset {?}{=}}\Lambda _{\bar {\mu }}{}^{\rho }\,\Lambda _{\bar {\nu }}{}^{\sigma }\,g_{\rho \sigma }.}$

as per discussion

http://www.physicsforums.com/showthread.php?p=3398120#post3398120 (especially post #36) JDoolin (talk) 14:51, 11 July 2011 (UTC)[reply]

I withdraw the question based on post #39 in the same thread. Thanks. (JDoolin (talk) 15:11, 12 July 2011 (UTC))[reply]

First, in the section "Definition", g is defined as a 4 x 4 metric tensor, but in the section "Local coordinates and matrix representations" g is defined as a scalar valued bilinear form (?)

${\displaystyle g=g_{\mu \nu }dx^{\mu }dx^{\nu }.\,}$

There is also, parenthetically, a third definition of g as a tensor field.

Finally, there is a definition of ds² as the line element and as the "metric", but the line element is ds, not ds².

I suggest separate, clear, correct and unambiguous definitions of the metric tensor, the metric, the tensor field, and the line element.

Then there should be a statement regarding the informal conflation of these by physicists who know what they are doing despite appearances to the contrary.

Perhaps there should also be an explanation of the relation of the element of proper time to the line element, i.e., dtau = ds/c.

200.83.113.147 (talk) 02:37, 19 February 2015 (UTC)[reply]

There is only one definition. These are merely showing the relationship between different notational schemes applied to the same notion of a metric. JRSpriggs (talk) 16:23, 20 February 2015 (UTC)[reply]

I found it confusing that g is introduced as the conventional notation for the metric tensor but later appears equated to a scalar. I believe that in such a basic article as this on the metric tensor that such confusion should be avoided, although I think it a good thing to point out that such confusing (to me) uses of terminology are quite common among physicists. Also, according to the article "Line element", the line element is ds not ds². 200.83.101.31 (talk) 02:55, 1 March 2015 (UTC)[reply]

It is requested that a physics diagram or diagrams be included in this article to improve its quality. Specific illustrations, plots or diagrams can be requested at the Graphic Lab. For more information, refer to discussion on this page and/or the listing at Wikipedia:Requested images.

This article is very math-heavy and could use some visualizations to be comprehensible to more readers. -- Beland (talk) 22:45, 13 November 2015 (UTC)[reply]

This article is mostly tensor analysis. You cant really draw tensors like vectors because they are multilinear mappings between vectors and/or dual vectors. Most of what can be visualized e.g. spherical coordinates can be found in the linked articles. Is there anything specific you want me to draw? Please say and I'll try. Thanks, M∧Ŝc²ħεИ_τlk 00:02, 14 November 2015 (UTC)[reply]

See Einstein field equation#The correspondence principle and show (expand) the derivation. There it is shown that

${\displaystyle g_{00}\approx -c^{2}-2\Phi \,}$

where Φ is the Newtonian gravitational potential (Joules per kilogram).

Also see Parameterized post-Newtonian formalism for a more detailed correspondence. There U is used for the Newtonian gravitational potential (except for a constant factor).

Furthermore, the Christoffel symbol (gravitational force field and basically the first derivative of the metric) is analogous to the electromagnetic field (basically the first derivative of the electromagnetic potential). JRSpriggs (talk) 11:15, 5 August 2017 (UTC)[reply]

Once again I find these math articles in Wikipedia just awful. Particularly: they don’t define terms. I want to know, for example, what is a metric tensor? Well, article starts by saying it’s the “fundamental object of study.” Really? That wasn’t my question. Then it goes on to say it’s like something in Newtonian physics. Geez! Didn’t we all hear our grammar school teachers telling us ”Johnny, don’t define a word by saying it’s like when you… that’s not a definition.” In every article the first sentence should be: “A metric tensor (or whatever) is…” Then say exactly what it is, not how it’s studied or how it resembles your aunt Edna. I suspect that mathematicians who write these articles are conversant with the computational processes involved, and can plug in variables and generate answers, but they don’t have a clear conception of the physical meaning of these terms. I do not accept the contention that mathematic terms cannot be expressed in words. I accept it’s difficult, but we look to encyclopedias to provide these answers. Our greatest physicist, Richard Feynman, also a great teacher, famously said: “if you cannot explain something such that a sophomore can understand it, it’s because you don’t understand it.”98.162.189.119 (talk) 20:51, 30 August 2021 (UTC)[reply]