Talk:Metric tensor

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Needs references and lead. Geometry guy 22:30, 17 June 2007 (UTC)

1 Request
2 Support for keeping this page
3 Is the first equation correct?
4 Trouble with non-metrics
5 Pseudo-Riemannian metrics
6 Riemannian metrics
7 Accessibility
8 intro reworking
- 8.1 v1
9 Relation to energy inner product
10 Non-L2 norms?
11 Vector transformation
12 How to calculate the matric tensor G?
13 clarity
14 accidental removal?
15 Question related to metric tensor...
16 Wow
17 (0,2) tensor

[edit] Request

Please do not delete this page. although there is an alternative approach to differential geometry, the component-based approach is fundamental to understanding the 'modern' approach, and the metric tensor is the fundamental definition of Riemannian geometry. What are the goals of this encyclopedia? what should they be? to be esoteric and create what some very few people might find to be 'elegant' and 'precise', or to make information accessible? I believe that it is the latter. Furthermore, I do not believe that the two goals are mutually exclusive. I believe, rather, that writting in a clear language that can readily be understood is a form of eloquence and 'perfection', and should be a priority. I am reminded of early medicine, when the professors turned the science of medicine into an esoteric and pedantic rite in pursuit of the luster of exclusive power. I would hate to see mathematics go the same way.

I like your attitude. I don't suppose you know what a tensor product is? See my comment on Talk:Tensor product. By the way, Kevin, you should sign your entries on talk with ~~~~, which is automatically replaced with a signature like the following. -- Tim Starling 04:23 Mar 14, 2003 (UTC)

perhaps we should explain the implicit summation and products of differentials more? - Gauge 05:41, 31 Jul 2004 (UTC)

[edit] Support for keeping this page

This page is simple, clear, and essentially self-contained. Browsing from the General Relativity entry, I was much happier with this page than with most other tensor-related explanations, which were so extravagantly reference-dependent as to be useless. I have a pretty solid general math and physics background, and doubt that a much more demanding presentation would serve a significant number of readers. Peter 19:50, 4 Feb 2005 (UTC)

I wonder if anybody has thought about the divide and conquer approach for writing mathematical objects like equations. It requires only the existence of parallel computers normally used in business (say stores).

Benjamin Cuong P. Nghiem bcnghiem@hotmail.com

Question:

Shouldn't be there a Sigma symbol in the first equation on this page? Under the square root. Something doesn't seem right without it in comparision to equation just under "The length of a curve reduces to the familiar calculus formula:" paragraph.

The summation convention is in force; so the Σ is implied. Charles Matthews 15:57, 4 August 2005 (UTC)

[edit] Is the first equation correct?

While I understand that the Einstein convention of summation over repeated indices is being employed in the first equation (so there is no need for the sigma); are those dt denominators within the square root right? 203.52.176.26 03:06, 6 December 2005 (UTC)

Yes, this equation is correct. Here is the reasoning: (x₁(t), ..., x_n(t)) is the equation describing the the curve in the local coordinate system, a ≤ t ≤ b. Assume that functions x_i(t), 1 ≤ i ≤ n are differentiable in the open interval a < t < b. Let us now calculate the distance, ds, between two infinitesimaly close points t and t + dt on this curve. If this curve were in n-dimensional Euclidean space, E_n then (using Einstein summation convention)

$ds = \sqrt{({dx_i \over dt} dt) ({dx_i \over dt} dt)} = \sqrt{{dx_i \over dt} {dx_i \over dt}} dt$ .

(This equation is generalization of Pythagoras theorem to n-dimensions). In a general Riemann space we have to add the metric tensor:

$ds = \sqrt{g_{ij} {dx_i \over dt} {dx_i \over dt}} dt$

And, finally, the length, L, of the curve segment between a and b is the integral of ds from a to b.

TomyDuby 07:56, 14 September 2007 (UTC)

[edit] Trouble with non-metrics

For the Minkiwsi case as well as for the Schwarzschild as well as for any "metric" from relativity theory, the metric tensor is not positive definite and especially the formula for the length of a curve does not apply in the form given in the article for a proper, positive definite metric simply because the root function is not well-defined for non-negative values and using the principal square root convention takes us just into some different kind of trouble.

So I've cut the calculation that leads to imaginary lengthes from the article and put it here:

" Indeed, the distance between A = (0,0,0,0) and B = (1,0,0,0) is

$L = \int_0^1 \sqrt{-(dx^0)^2} = \int_0^1 i dx^0 = i$ (with the principal square root convention, where $\sqrt{-1}$ is equal to i and not -i).

We can check that in this case, the distance from A to B is equal to the distance from B to A :

$L = \int_1^0 \sqrt{-(dx^0)^2} = \int_1^0 -i dx^0 = i$ (in this case, dx⁰ is negative).

The Infidel 19:37, 18 February 2006 (UTC)

The calculation as such (despite not done by me ;-) looks correct. Only the formula is wrong, but I havn't found a relayable reference with a correct one. The Infidel 17:24, 19 February 2006 (UTC)

[edit] Pseudo-Riemannian metrics

They are not "non-metrics"; they are pseudo-Riemannian. In the case of Minkowski and Schwarzschild, we have a particular type of pseudo-Riemannian: the "Lorentzian" metrics. These are all valid metrics. It is important that the general page on metric tensors includes all types of metric. I've changed the introduction to reflect these facts. I think I've also made it clearer.

I corrected a few flat-out errors. Also, because we can deal with metrics without coordinates, I took the coordinates out of the introduction. However, since coordinates are very useful in dealing with metrics, I just moved the coordinate expressions down a section. I think the page looks much better now, besides being more accurate. I hope you folks approve. MOBle 13:04, 22 February 2006 (UTC)

Not quite so. The bilinear forms that are not positive-semidefinite don't give rise to a mertric. On the other hand, physicist call the bilinear form a metric tensor without any regards of its signiture and often leave out the word "tensor" for convenience. I think this can be called separation by common language. The Infidel 18:40, 22 February 2006 (UTC)

[edit] Riemannian metrics

The term "Riemannian metric" now redirects to "Riemannian manifold". I think this will make curious math students happier. This way we can all be happy.

Tajmahall 07:08, 12 September 2006 (UTC)

[edit] Accessibility

I just read this article, but I still have a rather basic question; what the heck is a metric tensor? I understand that you have to give a precise definition for College math people and such, but could you put a working definition in layman's terms so that those of us that don't know much about college math can understand it? Thanks! Ahudson 18:50, 6 February 2007 (UTC)

Sorry, this article is in such terrible shape. It badly needs an overhaul. A metric tensor is an object defined on a manifold (like a sphere or torus or just ordinary 3-dimensional Euclidean space) that allows one to define geometric properties of that space; things like volume, lengths of curves, angles between curves and so on. Ultimately, it can tell you how lumpy, flat, or curved a space is. To be more precise, every point on a manifold has a vector space associated with it called the tangent space. A metric is something which defines a scalar product (dot product/inner product) on each tangent space. These scalar products aren't allowed to jump widely from point to point but must vary smoothly as one goes around the manifold. Moving from this basic definition to understanding geometric properties takes a lot of work and is the subject of Riemannian geometry. -- Fropuff 19:18, 6 February 2007 (UTC)

Ok, thanks! I think I get it now. Ahudson 16:47, 2 March 2007 (UTC)

[edit] intro reworking

[edit] v1

A metric tensor is an object defined on a manifold (like a sphere or torus or just 3-dimensional Euclidean space) that allows one to define geometric properties of that space, such as volume, lengths of curves, and angles between curves. It can also be used to determine how curved a space is at any given point.

Every point on a manifold has a vector space associated with it called the tangent space. A metric is something which defines a scalar product (dot product/inner product) on each tangent space. These scalar products vary smoothly as one goes around the manifold. Kevin Baas^talk 23:14, 3 March 2007 (UTC)

[edit] Relation to energy inner product

In functional analysis there's an energy inner product, written $\langle Bu|v\rangle$ , which appears to be the same sort of expression as $g_{ij}\,dx^i/dt\,dx^j/dt$ . That is, both are vector-matrix-vector products that yield a scalar. Is it accurate to say that the B in an energetic inner product is the metric tensor for that space? —Ben FrantzDale 13:05, 16 April 2007 (UTC)

kind of. one can be view these via a general construction of inner product spaces (see, for example, positive definite kernel). the simplest is an inner product on Rⁿ induced by a positive matrix. assigning a positive matrix smoothly to (the tangent space of) each point in a manifold gives a Riemannian metric. if your operator B, acting on some Hilbert space H, is bounded and > 0, then <x, Bx> is an inner product on H. the correponding norm "renorms" H. for example, if you do this in R², IIRC the unit circle in this new norm is some ellipse with axes parallel to the eigenvectors of B. the operator B you refer to seems to be unbounded in general, but the idea is the same, although i doubt it's called a "tensor" in functional analysis literature. Mct mht 01:32, 17 April 2007 (UTC)

[edit] Non-L2 norms?

It looks like there is an assumption that $\|dx\|=\sqrt{g_{ij} dx^i dx^j}$ . This is the case with a Hilbert space, but not with a Banach space in general. Could one have an equivalent of a metric tensor for norms involving other powers? For example, something like a max norm,

$\|dx\| := \max_{i} g_{ij} dx^j$

or a one-norm,

$\|dx\| := \sum_{i} g_{ij} dx^j.$

So, is this something that gets done, or is a metric tensor really only used for $L 2$ -like norms? —Ben FrantzDale 03:17, 27 April 2007 (UTC)

looks to me like the norms you have there won't be given by a tensor; there's something funny with those expressions (maybe the tensor and the Jacobian got mixed up?). but yes, there are non-Hilbert space norms. see Finsler manifold. Mct mht 04:05, 27 April 2007 (UTC)

Cool. That's exactly what I meant. I'll cross link this page with that one.

As for the norms I wrote above, what do you mean that they "won't be given by a tensor"? Is my nomenclature wrong? Certainly I could find the maximum or sum of

g i j d x j

, right? Is

g i j

not strictly-speaking a tensor in that usage? Thanks again. —Ben FrantzDale 04:24, 27 April 2007 (UTC)

looks like they would violate the axioms of a norm, say positivity.

what i meant was that, locally you have a positive definite matrix G > 0 that induces an inner product on the tangent space V. denoted by (V <, >_G). let J be the Jacobian, so G = J^TJ, and Ran(J) be its range, then (V <,>_G) is really isomorphic, in the sense of inner product spaces, to (Ran(J), <, >) where <, > is the usual Euclidean inner product. so abandoning <,>_G on V is abandoning the usual <,> on Ran(J). and to introduce a new norm, perhaps one should be talking about J. for example, the ∞-norm on Ran(J) would be

$\|dx\| := \max_{i} | J_{ij} dx^j |.$

maybe someone would be willing clarify for us these issues. Mct mht 04:56, 27 April 2007 (UTC)

That makes sense. I forgot to include absolute value in there. Thanks. —Ben FrantzDale 06:23, 27 April 2007 (UTC)

[edit] Vector transformation

I'm not sure if I should post this here or in Talk:Inner product space. When I see $g i j x i y j$ I want to think about it as transforming x and y and then taking their good-old-fashioned inner product. But it seems like that transformation, assuming g is self-adjoint, is going to be $\sqrt{g}$ . That is, consider

$\begin{bmatrix}1& 0\end{bmatrix}\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}=2$ .

If I wanted to transform the vectors separately first, I would have to do

$\begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}\sqrt{2} & 0 \\ 0 & \sqrt{2}\end{bmatrix} \begin{bmatrix}\sqrt{2} & 0 \\ 0 & \sqrt{2}\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}\sqrt{2} & 0\end{bmatrix}\begin{bmatrix}\sqrt{2}\\0\end{bmatrix}=2$ .

In other words, the inner product space you get by using a metric tensor, g, is the same as what you'd get by transforming all vectors by $\sqrt{g}$ . Is that a reasonable way of thinking about it? —Ben FrantzDale 06:17, 30 April 2007 (UTC)

essentially, yes. G is > 0, so G = B*B for some invertible B. there's a natural isomorphism between (V, <,>_G) and (Ran(B), <,>). take v in (V, <,>_G) to Bv in (Ran(B), <,>). this map preserves the inner product and so is an isomorphism. replace B by the positive square root G^1/2 and we get something that seems pretty close to what you're saying.

same thing goes through when G is ≥ 0. but now <,>_G may be degenerate and Ran(B) may be a proper subspace. the map is then [v] -> Bv, where [v] is an equivalence class modulo the degenerate subspace. Mct mht 06:41, 30 April 2007 (UTC)

[edit] How to calculate the matric tensor G?

I miss a simple explanation how to do this. TomyDuby 07:58, 14 September 2007 (UTC)

To compute the metric tensor from a set of equations relating the space to cartesian space (g_ij = δ_ij: see Kronecker delta for more details), compute the jacobian of the set of equations, and multiply (outer product) the transpose of that jacobian by the jacobian.

G = J T J

Was that helpful? Kevin Baas^talk 01:59, 17 September 2007 (UTC)

Yes, it is helpful.

Thanks. TomyDuby 21:07, 21 September 2007 (UTC)

[edit] clarity

In 4 years,

"To compute the metric tensor from a set of equations relating the space to cartesian space (gij = δij: see Kronecker delta for more details), compute the jacobian of the set of equations, and multiply (outer product) the transpose of that jacobian by the jacobian."

G = J^T J

has become:

The induced metric tensor for a smooth embedding of a manifold into Euclidean space can be computed by the formula

G = J^T J

where J denotes the Jacobian of the embedding and J^T its transpose.

The following was inserted into the intro, as if to clarify:

In other terms, given a smooth manifold, we make a choice of positive-definite quadratic form on the manifold's tangent spaces which varies smoothly from point to point. The manifold, equipped with the metric tensor (the varying choice of quadratic form), is called a Riemannian manifold and in this context the metric tensor is often called a Riemannian metric.

And everything from "Arbitrary Coordinates" on down was added to the article.

Now I have nothing to say about anything below "arbitrary coordinates", but i take issue with the other two things: I don't see them as improvements.

The first part: most people reading the article have no idea what an "induced" metric tensor is, what a "smooth embedding" is, why it is relevant, nor what "euclidean space" is. Or how to interpret the formula given. Though i'm sure many a reader comes to the article with a set of things they easily recognize as "equations" that relate the space to "cartesian space", a concept they've very familiar with, and want to know how to compute the "metric tensor" from them. To those people, the new version of the paragraph isn't much help.

As to the second part: I can tell right away that english isn't the author's first language. And again, people shouldn't be expected to know what a "form" is, what a "tangent space" is, etc. And if I were to take a poll of how many people those two sentences makes any sense to, I know that I would get a very, very, very small set of hands. I know I wouldn't be raising mine.

I'm quite compelled to revert the first part back four years, and remove the second part. Kevin Baas^talk 01:58, 17 September 2007 (UTC)

i am going to revert the last 4 edits (adding back the passage and removing one wrong edit). i didn't write that passage but it seems fine and is an accurate desccription of what the metric tensor is, as opposed to merely how it's calculated. saying the author's first language doesn't seem to be english is an inexplicable cheapshot and just bs. not understanding something ain't a valid reason to chuck it in this case. Mct mht 23:59, 19 September 2007 (UTC)

No, I wasn't making a cheapshot, I being serious. It honestly looks to me like the author's first language isn't english. The mistakes in the construction of the sentences were those that are commonly made when someone is speaking/writting in a language that is not their native language:

Miswording or not knowing idioms: "In other terms" should be "In other words".
Redundancy that adds confusion showing a poor understanding of proper sentence syntax: "smooth manifold...which varies smoothly from point to point" - using proper english rules for the interpretation of syntax, this tells the reader that there is something in the sentence that varies smoothly from point to point, but it is not the "smooth manifold". However, the sentence does not make clear what that noun is.

Additionally:

the "In other terms (sic)..." sentence is NOT a restatement of the sentence that comes before it; i.e. it is NOT "in other words". again, confusing the reader.
Each sentence has more than 7 ideas in it, thus breaking the rules of good writing style. I'd guess that the author's native language is German, as I've heard their sentence syntax is pretty loose and their sentences often relatively complex.

Kevin Baas^talk 22:31, 20 September 2007 (UTC)

Oh, and I put my minor corrections back. I assume your reverting them was a mistake, as their pretty non-controversial edits. But I left the disputed text at the status quo so that it can be discussed peacefully. Kevin Baas^talk 22:34, 20 September 2007 (UTC)

[edit] accidental removal?

something funny happened in [this edit], especially considering it was Michael Hardy. material got removed that shouldn't have been. i am gonna leave a message on his talk page. Mct mht 00:22, 20 September 2007 (UTC)

[edit] Question related to metric tensor...

How do the dot products of covariant basis vectors produce a contravariant component of the metric tensor and vice versa? I.e why do the variance types switch through taking the dot product of basis vectors? I don't see it coming out in the math. Thanks.

If I understand your question, you're misusing "dot product". A dot product takes two vectors and gives you a scalar (which has no variance type). Rather, I think you mean to talk about taking the contraction of the basis vector with the metric.

In index notation, we could write this as

g μν v ν = v μ

, where

v ν

is the covariant form of the basis vector. Then,

v ν

is the contravariant form. Alternatively, we could get the covariant form from the contravariant form:

g μν v ν = v μ

Now note that, because of how the metric components transform, you get two extra factors of the Jacobian or inverse Jacobian, which is why the variance type switches. That's not hard to check. (Note that some times it's confusing to have indices on basis vectors to indicate which basis vector you're dealing with, so I've left them off. Is that where your confusion was coming from?) --131.215.123.98 (talk) 17:32, 25 November 2007 (UTC)

Thanks for replying. I think you might be answering my question, but it's going a few steps over my head. Let's back up a bit, and I'll first make more clear the context with which I'm using to approach this metric tensor concept and where my confusion is arising. Specifically, I'm reading this article http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/documents/Tensors_TM2002211716.pdf He is saying there that the dot products of any two of the bases in one of the variances gives you a corresponding component in the metric tensor (of the opposite variance). It is at the bottom of page 20 and I can't see where this conclusion comes from based on any of the previous material. It is the sole point in the whole article I can't wrap my head around. Cheers. —Preceding unsigned comment added by 160.39.130.184 (talk) 03:24, 26 November 2007 (UTC)

[edit] Wow

This article is breathtakingly bad. It lacks the one thing it should without question have: a clear mathematical definition of what a metric tensor is. Namely, a section of the tensor product of the cotangent bundle of M with itself, which at each point gives us a positive-definite symmetric bilinear form (i.e. an inner product). See e.g. http://planetmath.org/encyclopedia/MetricTensor.html. The introductory paragraph should not attempt a precise mathematical definition, rather giving the intuition of the idea. But later on, at least a paragraph needs to be devoted to exactly what a metric tensor is, why it's of type (0, 2), how we get from g to a matrix using coordinates, etc. And this should probably be the first body paragraph, before talking about applications -- how to calculate things with metrics and so on. I would also note that the article as it stands does things in coordinates wherever it can, which is a remarkably bad idea. Sometimes coordinates are necessary, but it's better, more canonical, less cluttered, easier to understand, to do things coordinate-free when possible. For example, instead of saying the cosine of an angle is g_iju^iv^j/sqrt(g_iju^iu^j*gijv^iv^j) (by the way, the absolute values are not needed), why not say it's g(u, v)/sqrt(g(u, u)*g(v, v))? I would note that this article used to have a "definition" section, some months ago, which Michael Hardy for some reason removed. Can't exactly fathom why that would be. Somebody want to fix this article? If there are no takers, I'll try and have a go at it, but I don't really have much time to devote to it and it wouldn't get done any time soon. Kier07 (talk) 16:56, 9 May 2008 (UTC)

Go for it. —Ben FrantzDale (talk) 02:13, 10 May 2008 (UTC)

Why not go back to the pre-"Michael Hardy" point and use that as a new starting point?RandomTool2 (talk) 16:45, 10 May 2008 (UTC)

[edit] (0,2) tensor

what is a (0,2) tensor? This should be explained or have a wikilink. Thanks. 131.111.28.28 (talk) 12:50, 20 May 2008 (UTC)

While I have never encountered this (0,2) terminology, I would assume that it refers simply to the fact that the (covariant) tensor has 2 lower, and 0 upper indicesRandomTool2 (talk) 19:08, 1 June 2008 (UTC)