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From Wikipedia:

From the coordinate-independent point of view, a metric tensor field is defined to be a nondegenerate symmetric bilinear form on each tangent space that varies smoothly from point to point.

I'm not sure I truly get this. The metric is locally a flat Lorentzian metric. We can write it in a variety of coordinates (and use a standard measure stick). But if a manifold is curved we have to include variations of the metric to say if the manifold is curved. But if it's locally flat how then can it encode for curvature (if it contains this information in the first place, of which I'm not sure)?

So, is the metric an intrinsic property of spacetime, independent of the coordinates, or is it coordinate dependent, the manifold inducing changes in it while moving from point to point? But if so, how can changes be induced if it's locally flat at every point?

EDIT

I understand that the flatness is approximate. But how does this relate to the metric if you move from point to point if the metric at a point is defined by looking at an infinitesimally small region around the point in question? If we change coordinates this metric doesn't change, although the expression in coordinates of course does. Like the length of a vector doesn't change if you use different coordinates.

So let me ask it differently. How can we write a coordinate-free expression for the metric? Is it just the same as for a vector? So the square of the components is always a fixed "length"? Is this what is meant in the definition above (from Wikipedia)?

To put it differently, can we consider the metric field somewhat like a vector field with the lengths of the vectors varying in the field?

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    $\begingroup$ You make invariants from a tensor like the metric $g_{\mu\nu}$ the same way you do for a vector $v^\mu$: find suitable things to contract it with until there are no more indices. $v_\mu v^\mu$ is the squared length of a vector and e.g. $D=g_{\mu\nu}g^{\mu\nu}$ is the dimension of spacetime. But a) fully "coordinate free" notations, like $D=\mathrm{tr}(\mathbf g\mathbf g^{-1})$ are considered unwieldy as the number of things you can do with tensors grows with rank faster than you can invent notation and b) the invariants of a vector or tensor are not enough to determine the original object. $\endgroup$
    – HTNW
    Aug 1, 2022 at 19:03

3 Answers 3

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It is an intrinsic, coordinate-independent property.

While coordinate changes allow you to get the metric at a point to be the Minkowski metric and the Christoffel symbols to vanish at that point, no coordinate change allows you to get the Riemann tensor to vanish in a curved manifold. In other worlds, changing coordinates let's you pick the metric to be the Minkowski metric (at a point) and let's you pick it's derivatives to be zero (at the same point), but you can't make the second derivatives vanish.

About the Edit

How can we write a coordinate free expression for the metric? Is it just the same as for a vector?

I never saw one, and I don't really think one can write a completely coordinate-free expression for a specific metric. This is similar to how you always need to use a coordinate system to write a vector. For example, $\mathbf{v} = 2 \mathbf{\hat{x}} + 3 2 \mathbf{\hat{y}}$ is a vector and it is an object which does not depend on choices of basis or coordinates. Nevertheless, I had to choose a particular basis to write it down. The basis for a metric tensor is slightly more complicated: at a given point in the manifold, it is the basis of the tensor product $V^* \otimes V^*$, where $V$ is the tangent space at that point (we can also talk about the basis in terms of tensor fields, but that will introduce an extra level of mathematical complication without much insight gain, in my opinion).

So the square of the components is always a fixed "length"?

The expression for the length might not be just the square root of the sum of the squares of the components. As an example, the length of a vector in Euclidean space is not given in this way when one employs spherical coordinates.

Is this what is meant in the definition above (from Wikipedia)?

What is meant in the definition of Wikipedia is that the metric (and hence lengths and angles) does not depend on coordinates, even though it is convenient to use coordinates when expressing them. The expression for the lengths and angles might change, because the components of the metric might change.

To put it differently, can we consider the metric field somewhat like a vector-field with the lengths of the vectors varying in the field?

At each point, the metric is a tensor, which is just an element of the vector space $V^* \otimes V^*$ (in the case of a metric). Hence, in this sense, the metric is a vector (notice I mean this in the mathematical sense of the world: the metric is not a covariant nor a contravariant vector).

About the Comments

The metric already encodes all of the information about curvature, which can be obtained by differentiating the metric: from the metric one computes the Christoffel symbols, from which one can compute the Riemann tensor. At the end of the day, the Riemann tensor is built with the zeroth, first, and second derivatives of the metric. These definitions are fully general and apply to any sufficiently differentiable spacetimes (including black holes) and to Differential Geometry in general.

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  • $\begingroup$ So the information about the curvature is encoded in the metric already? And the metric is an intrinsic (to the manifold) coordinate independent tensor? If we "roll" the flat tangent space (or the manifold over a flat plane) over the manifold It's the second derivative that counts? Which is non-zero even if the first derivative is zero? How does this work out for a black hole? How does the true manifold look if in one coordinate system it looks like a manifold where the radial coordinate changes into time behind the horizon while in the falling frame there is no horizon at all? $\endgroup$
    – Gerald
    Aug 1, 2022 at 15:55
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    $\begingroup$ @Gerald That's a lot of questions at once, especially in a comment rather than a question. The metric tensor and its first two derivatives all matter to the Riemann tensor, and only if all of that tensor's components vanish is the spacetime flat. What vanishes where, in terms of first derivatives at most, depends on the coordinates chosen. If you have specific concerns about any black hole metric (e.g. Schwarzschild or Kerr), I recommend asking a new question. $\endgroup$
    – J.G.
    Aug 1, 2022 at 16:05
  • $\begingroup$ @J.G. Thanks for your comment. I accidentally pushed the red flag... But could pull it down again. $\endgroup$
    – Gerald
    Aug 1, 2022 at 16:59
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    $\begingroup$ @Gerald I updated my answer to address some of the points you raised in the edit and the comments, but I restricted myself to the points asking for further clarification on this very question. As mentioned by J.G., avoid asking more than one question in a single post, unless they are extremely related. I tried to answer all the follow-ups that are necessary to understand the answer to the original question, but if you'd like further clarification on some point related to this question please let me know =) $\endgroup$ Aug 1, 2022 at 17:14
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    $\begingroup$ @NíckolasAlves Thanks for your clear answer! I wait a bit accepting it. $\endgroup$
    – Gerald
    Aug 1, 2022 at 20:24
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Nickolas' answer is excellent, so I'd just like to supplement it with an example of a coordinate-independent definition of a metric on the 2-sphere, which I will define as $$\mathrm S^2:= \{(x,y,z)\in \mathbb R^3 \ | \ x^2+y^2+z^2=1\}$$

Note that a 3-tuple $(x,y,z)$ is not a list of coordinates of a point $p\in \mathrm S^2$; it is an actual element of the set underlying $\mathrm S^2$. Being two-dimensional, $\mathrm S^2$ is locally homeomorphic to $\mathbb R^2$, not $\mathbb R^3$, and points in $\mathrm S^2$ have two coordinates, not three.

Define the subset $U\subset \mathrm S^2$ as follows: $$U:=\left\{(x,y,z)\in \mathrm S^2 \ \bigg|\ y \neq 0 \ \vee\ x<0 \right\}$$ enter image description here On this domain, define the following functions: $$\theta: (x,y,z) \mapsto \cos^{-1}(z) \in (0,\pi)$$ $$\phi:(x,y,z) \mapsto \mathrm{atan}(x,y)\in (0,2\pi)$$ Finally, define the metric $$g = \mathrm d\theta \otimes \mathrm d\theta + \sin^2(\theta) \mathrm d\phi \otimes \mathrm d\phi \tag{$\star$}$$ which acts on two vectors $X,Y\in T_p M$ to yields $$g(X,Y) = \bigg[X(\theta)Y(\theta) + \sin^2\big(\theta\big) X(\phi) Y(\phi)\bigg]_p$$

Nowhere in the above description have I used any coordinates. Of course, $\theta$ and $\phi$ are perfectly good coordinate maps on the domain $U$.


As an alternative construction, note that any curve $\gamma:\mathbb R\rightarrow \mathrm S^2$ trivially defines curve in $\mathbb R^3$ via the inclusion map $\iota :\mathrm S^2 \hookrightarrow \mathbb R^3$. As a result, any vector $X$ on $\mathrm S^2$ can be pushed forward to a vector $\iota_*X$ on $\mathbb R^3$. Given a metric $g$ on $\mathbb R^3$, we may define the pullback metric $\iota^*g$ on $\mathrm S^2$ which is defined as $$\big(\iota^*g\big)(X,Y) = g(\iota_*X,\iota_*Y)$$

In particular, if we define $x:(a,b,c)\mapsto a$, $y:(a,b,c)\mapsto b$, and $z:(a,b,c)\mapsto c$ to be smooth functions on $\mathbb R^3$ and then take the Euclidean metric $g = \mathrm dx\otimes \mathrm dx + \mathrm dy \otimes \mathrm dy + \mathrm dz \otimes \mathrm dz$, then $\iota^*g$ coincides with the metric defined in $(\star)$.

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To implement the "local flatness" condition, you pick a point, set the coordinates at that point to zero. It is then possible to scale and direct the axes off that point in such a way that the metric $g_{μν}(x)$ at $x = 0$ has the form $$g_{μν}(0) = η_{μν},\quad ∂_ρ g_{μν}(0) = 0,\quad A_{νμρσ} = A_{μνρσ} = A_{μνσρ}.$$ Then, by Taylor's Theorem, you can write, in the vicinity of $x = 0$: $$g_{μν}(x) = η_{μν} + \frac12 A_{μνρσ} x^ρ x^σ + O\left(x^3\right).$$ The second order coefficients are directly related to the components of the Riemannian curvature tensor at $x = 0$. You may want to substitute this into the expression for the curvature tensor and evaluate it at $x = 0$ to see just what their relation is.

So, you asked how a locally-flat metric can produce a curvature tensor. That's how. It's in the second order derivatives.

There's even a closed-form expression for a full Taylor expansion of the metric in terms of the Riemann tensor and its derivatives at $x = 0$, which I don't recall the reference anymore; but ... the Internet does: A Closed Formula for the Riemann Normal Coordinate Expansion. In the future, when people do cites, they'll be like Mathematicians and instead of saying what or where it is, they'll just write down "it exists" and that'll count as the citation.

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