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Does the metric tensor only depend on the coordinate system used?

  1. May 24, 2012 #1
    I have looked at the definition of the metric tensor, and my sources state that to calculate it, one must first calculate the components of the position vector and compute it's Jacobian. The metric tensor is then the transpose of the Jacobian multiplied by the Jacobian.

    My problem with this is the way the Riemann Curvature Tensor is defined. For a given manifold, it can be calculated (the curvature tensor) by using the metric tensor and Christoffel Symbols. But if it only requires those, then how does it in any way describe the curvature of the manifold as the metric tensor is according to the aforementioned definition dependent only on the employed coordinate system.

    Thank you in advance for any help you may provide.

    I have attempted to calculate the metric tensor using the Jacobian method (mentioned above) and it has worked (I tried it on Cartesian, Polar and Spherical coordinates).
     
    Last edited: May 24, 2012
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  3. May 24, 2012 #2
    Just to rephrase what he is asking (which I also want to know):

    The question is how is the metric tensor calculated given a certain manifold. I.e., how are the individual components, which are given by derivatives of coordinates with respect to other coordinates calculated given a certain manifold? For example, I can imagine the manifold being a spehere are some other shape. How are the derivatives then calculated.

    Btw HilbertSpace, I think that the metric tensors you have calculated are done with different sets of coordinates, but all in Euclidean Space, am I right?
     
  4. May 24, 2012 #3
    Yes, Guillefix. I used different coordinate systems (ergo the position vector had different components when using i,j and k unit vectors) but all in Euclidean Space.
     
  5. May 24, 2012 #4
    Hello.

    1. If you change coordinates, the components of the Riemannian curvature tensor change along with the metric tensor components. It might help you to think of it like this: The metric is actually an inner product g(X,Y) which is well defined for any vectors X and Y tangent to the manifold at point p. As you change coordinates, the components of X and Y change according to your coordinate change. Therefore, the coefficients of g must also change to compensate.

    The curvature tensor is the same, except that it is a 4-tensor R(X,Y,Z,W). So if you change coordinates, then the components for each vector change according to the Jacobian, therefore, so must the components of R. If your components for R did not change, that is when you would have something that does not make sense on the manifold.

    2. Guillefix, your question seems a bit different. Lets say you are talking about a submanifold of Euclidean space such as the sphere, which you mentioned. In that case, the first thing you need is a parametrization. Let u = colatitude, v = longitude (standard spherical coordinates).

    So the sphere is the image of the mapping (x(u,v), y(u,v), z(u,v)). This is a submanifold of Euclidean space where the metric is dx^2+dy^2+dz^2. So calculate the differentials dx, dy, dz, and just do regular algebra to get a formula E du^2+2F dudv+ Gdv^2. E,F,G are the components of the metric in local coordinates. You can also think of this as pulling the metric of the sphere back to the u,v plane. But it amounts to the same thing.

    Does that answer the questions?
     
  6. May 24, 2012 #5
    Thank you Vargo, I think I fully grasp the concept of the metric tensor now. However I still have some concerns with regards to the curvature tensor, but I think they will dissipate once I've done some problems (and begin to compute them). Thank you very much.
     
  7. May 24, 2012 #6
    Happy to help
     
  8. May 24, 2012 #7

    Fredrik

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    I don't think the question really makes sense, because a metric is something you add on top of the manifold structure. Just like a vector space can have several different norms or inner products, a manifold can have several different metrics. So the metric isn't determined from the manifold.

    So I'm guessing that what you have in mind are manifolds that are defined as subsets of ℝn, and endowed with the metrics that are induced on them by the Euclidean metric on ℝn. Denote the subset by M and the metric on ℝn by g. The "natural" metric on M is the pullback metric I*g, where I is the inclusion function ##I:M\to\mathbb R^n##, defined by I(x)=x for all ##x\in M##.

    Hm, I must have explained what a pullback is somewhere. Let's see if I can find a post like that...OK, this one will do: https://www.physicsforums.com/showpost.php?p=2534439&postcount=9
     
  9. May 27, 2012 #8
    Ok, i read a bit more, and got a bit more clear i think. Fredrik, what you mean that the metric isnt defined by the manifold is because the metric of a 2d manifold could be many different things depending on the surface it represents on a 3d manifold say? And other thing, can't a manifold always be expressed as a subset of a greater dimensional space? If so this is what I meant when i said how do you get the metric tensor from the manifold, i was imagining a hypersurface in a Euclidean space. I dont really get thr notation you use for the pullback function because its probably yet above my level, but i think i get it conceptually, is it just a function that using the metric in the ambient space, it asigns a metric to a certain hypersurface or other subspace in that space?
     
  10. May 27, 2012 #9

    Hurkyl

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    I don't know if it will help, but it may be interesting to see the complete characterization of the one-dimensional case.

    There are only two one-dimensional manifolds: "line" and "circle".

    There are many more one-dimensional Riemann manifolds:

    • Open line segment with length x
    • Ray
    • Line
    • Circle with perimeter x
    (x can be any positive real number)

    The types in the first three bullet points are metric structures that can be put on "line", and the last type are metric structures that can be put on "circle".

    It might look like there are many more one-dimensional manifolds -- e.g. every positive real-valued function on R yields a metric tensor on R. But the neat thing is that, for example, if you have two metric tensors that would give R infinite length in both directions, then there is a change of coordinates that transforms one into the other.
     
  11. May 27, 2012 #10

    Fredrik

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    Are you familiar with inner product spaces? Do you understand that many different inner products can be defined on the same vector space? A metric is almost the same thing as a choice of an inner product on the tangent space at p, for each p in the manifold. And it's certainly possible to define more than one inner product on each tangent space. That's why I suspected that you were talking about hypersurfaces that inherit their metrics from the Euclidean space that they're a subset of.

    Yes, there are theorems about how manifolds can be embedded in Euclidan spaces, but I don't know exactly what they say. In particular, I don't know what they say about the metric. I just did a quick search, and they're not talking about hypersurfaces (n-1-dimensional submanifolds of n-dimensional manifolds) of Euclidean space. You may need 2n dimensions for the embedding of an n-dimensional manifold, and I think that's without even considering the metric. If you want the metric to be inherited from the Euclidean metric, things seem to be more complicated. This discussion (with contributions from PF member George Jones) mentions that you may need 90 (or 91) dimensions for the isometric embedding of a Lorentzian 3+1-dimensional manifold.

    If you look at the definition of "manifold" and "metric", you will see that the metric is just a (2,0) tensor field on the manifold. There is certainly more than one of those.

    The concept of "pullback" isn't hard to understand when you understand the basics well, in particular the definitions of the tangent and cotangent spaces. If you want, you can ask about the notation that you find confusing.
     
  12. May 27, 2012 #11

    HallsofIvy

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    I would be inclined to say that not only is it not correct to say that the metric tensor "only depends on the coordinate system used" but to add that the metric tensor, like any tensor, is independent of the coordinate system used! That is, after all, the whole point of a "tensor". Its components may depend upon the coordinate system but the tensor itself does not. Just as a vector can be written as <1, 0, 0> or <0, 1, 0> depending upon how you set us the coordinate system but the vector itself is NOT those numbers.
     
  13. May 28, 2012 #12
    Yes, I agree, the metric conceptually defines length and area, which don't change with a change in coordinates

    Wow, 91 dimensions. I thought that just as any surface can be embedded in 3 dimensions, any n-hypersurface could be embedded in a n+1 space! I mean a 3 dimensional hypersurface may be something like t=f(u,v,w) x=u y=v z=w, isn't that in four dimensions? Or am I imagining manifolds wrong?

    For the case in which the metric is taken into account, i can only imagine one case for which inheriting it from the ambient space wouldn't work, altho im not sure: a projection of a hypersurface into a lower dimensional hypersurface. Then I suspect that the metric of the projection can't be inherited from its ambient space? I suspect this because information is lost with projection into lower dimensions.
     
  14. May 28, 2012 #13

    Hurkyl

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    A hypersurface is a very specific idea. Specifically, the phrase "X is a hypersurface of Y" means that Y is a manifold, X is a submanifold of Y, and the dimension of Y is one more than the dimension of X.

    The phrase "X is a hypersurface" is nonsense. It is only meaningful in a context where there is an implicit value for Y, so that "X is a hypersurface" is meant as shorthand for "X is a hypersurface of Y".
     
  15. May 28, 2012 #14
    Yes when I mention a hypersurface, I mean a hypersurface in general, i.e. Y can be anything. OK, in my post I was refering only to euclidean Ys
     
    Last edited: May 28, 2012
  16. May 28, 2012 #15
    Hmm, ok I think I get it. You can have 2-dimensional surfaces on 4-dimensions, say, and maybe you can't express that surface on 3D? Thus you can't embed the 2-manifold on 3D.

    According to Nash embedding theorem, the euclidean space on which the manifold can be embedded has n ≤ m(3m+11)/2 if M is a compact manifold, or n ≤ m(m+1)(3m+11)/2 if M is a non-compact manifold
    http://en.wikipedia.org/wiki/Nash_embedding_theorem

    I certainly have learnt something new today.
     
  17. May 28, 2012 #16

    HallsofIvy

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    That depends upon the manifold. You cannot, for example, embed the surface of the Klein bottle, a two dimensional surface, in Euclidean three space. But you can embed it in Euclidean four dimensional space and there exist a three dimensional subspace of that containing the surface of the Klein bottle.

     
  18. May 28, 2012 #17
    And it is not euclidean. So i guess that all manifolds are hypersurfaces of some manifold, wether euclidean or not?
     
  19. May 28, 2012 #18

    lavinia

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    - There is no three dimensional subspace of R^4 that contains the Klein bottle.

    - Another surface that can be embedded in R^4 but not in R^3 is the Projective plane.

    - Try proving that a closed surface in R^3 must be orientable. It follows from this that neither the Klein bottle nor the Projective plane can be embedded in R^3.
     
    Last edited: May 28, 2012
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