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I Riemann tensor in 3d Cartesian coordinates

  1. Feb 10, 2017 #1
    Suppose we wish to use Cartesian coordinates for points on the surface of a sphere. Then all derivatives of the metric would vanish and so the Riemann curvature tensor would vanish. But it would give us a wrong result, namely that the space is not curved. So it means that if we want to get correct results we must necessarily use spherical coordinates in this case?
     
    Last edited: Feb 10, 2017
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  3. Feb 10, 2017 #2

    Orodruin

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    You are using Cartesian coordinates in an embedding space. This is not the same thing as describing the intrinsic properties of the surface and even if the metric in the embedding space is flat, the induced metric on the submanifold can be curved.
     
  4. Feb 10, 2017 #3

    Nugatory

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    First you have to precise about what you mean by "using Cartesian coordinates on the surface of a sphere". If you're just attaching ordinary x,y and z coordinates to points on the surface of a soccer ball on the table in front of you, then you're working in a three-dimensional flat space (the room you're in) that happens to contain a soccer ball. A straight line between any two points on the surface of the ball will pass through the interior of the ball; a two dimensional creature living on the two dimensional surface of the ball would interpret that straight line as a path through the mystical third dimension that connects the two points without going through any of the points in between.

    The two-dimensional surface of the ball is curved, and it will be identically curved no matter what two-dimensional coordinates you use to label the points on its surface. Latitude and longitude are familiar and easy to work with, but there are others. Note that in this two dimensional world, the paths through the interior of the ball do not exist; a straight line between any two points is a great circle along the surface.
     
  5. Feb 10, 2017 #4
    Thanks Nugatory and Orodruin for your reply.

    I see I think. Good example Nugatory.

    So to get non zero components of the Riemann tensor we would have to make a displacement on the surface of the sphere? Using spherical coordinates it's easy, because one just have to set the radius constant. Using Cartesian coordinates I think it is more difficult to keep on a "straight line", because we would have to change three parametres, namely x, y and z. In this case, we would get non-zero derivatives of the metric. Is this right?
     
  6. Feb 10, 2017 #5

    Nugatory

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    The surface of the sphere is a two-dimensional surface, so you only get two coordinates (if you introduce a third, you are working with a three-dimensional space of which the points making up the surface of your sphere is an arbitrary subset). No matter what coordinates you use, you will find that there is no two-dimensional coordinate system in which the metric components are ##g_{ij}=\delta_{ij}## everywhere on the surface of the sphere - which is to say that surface is not flat and it cannot be properly described with two-dimensional Cartesian coordinates.

    Of course "flat" can be a pretty good approximation, and then we can use Cartesian coordinates locally. For example, if you ask a Manhattanite for directions, you're likely to hear something like "three blocks uptown, two blocks crosstown" - that's Cartesian coordinates with axes aligned along the street grid. But formally what we're doing is defining a completely different two-dimensional surface, namely a plane that happens to be tangent to the surface of the earth at the point that we're standing and using Cartesian coordinates to give directions for motion in that plane, not on the surface of the earth.
     
  7. Feb 10, 2017 #6
    Could it be accepted as a definition of Cartesian coordinates? That is, Cartesian coordinates are the ones which describe flat spaces, no matter what their dimension is?
    I see.
     
  8. Feb 10, 2017 #7

    Orodruin

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    No. Cartesian coordinates are coordinates on a Euclidean space. There is no necessity for a flat space to be Euclidean.
     
  9. Feb 10, 2017 #8
    Can you give me an example of a flat, non-Euclidean space in which we cannot use Cartesian coordinates?
     
  10. Feb 10, 2017 #9

    Orodruin

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    A flat torus, a flat Möbius strip, a cylinder, a flat Klein bottle, a cone with the apex removed, etc.
     
  11. Feb 10, 2017 #10
    In what sense a cylinder is flat?
     
  12. Feb 10, 2017 #11

    Orodruin

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    The usual sense. A cylinder embedded in R3 has zero intrinsic curvature. You can cut it open and lay it flat on a table.
     
  13. Feb 10, 2017 #12

    Ibix

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    A cylinder has no intrinsic curvature. Draw a triangle on a piece of paper and roll it into a cylinder. The angles still add to 180. Circles still have circumference ##2\pi r## if the radius is measured in the surface of the cylinder. Etcetera. It's pretty much the canonical simple example of a space with extrinsic curvature but no intrinsic curvature.
     
  14. Feb 10, 2017 #13

    Orodruin

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    Also, in general, a metric need not be induced by an embedding.
     
  15. Feb 10, 2017 #14

    DrGreg

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    You can tightly wrap a flat piece of paper round it without stretching, tearing or crumpling. So its "intrinsic" local structure (as a manifold) is no different than a flat piece of paper's. Of course they have different "extrinsic" properties (which aren't part of the manifold, but of the embedding in 3D Euclidean space) and different global properties, but flatness is a local intrinsic property.
     
  16. Feb 10, 2017 #15

    Nugatory

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    Take a sheet of ordinary paper. It will pass any test for local flatness you try: the interior angles of triangles will add to 180 degrees, the pythagorean theorem works, there is a coordinate system in which the metric components are ##g_{ij}=\delta_{ij}##. Roll the sheet up into tube and it will form the surface of a cylinder, but it will still have all those flatness properties. The easiest way to see this is to imagine that you drew some geometric shape on the piece of paper before you rolled it up - the marks on the sheet of paper don't move around on the surface so their geometric relationships don't change.

    The surface of a sphere is fundamentally different because there's no way of forming the sheet of paper into a sphere without stretching the paper (which doesn't work because paper doesn't stretch - better to use a rubber sheet instead) and altering the geometric relationships between nearby points.

    This would be a good time to google search for the difference between "intrinsic curvature" and "extrinsic curvature" if you are not already familiar with those terms.
     
  17. Feb 10, 2017 #16

    Orodruin

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    It is official. This thread is now a test of how many different ways one can say "make a paper roll". :rolleyes:
     
  18. Feb 10, 2017 #17
    @Orodruin , @DrGreg , @Nugatory and @Ibix, thanks for explaining what is the meaning of intrinsic and extrinsic curvature.

    Trying to understand the meaning of this, I have tried the following:

    Suppose we want to calculate the distance between two points on a sphere, where one point is on the pole and the other is on the equator of the sphere. We can set our coordinate axes such that one coordinate, say z, is always zero. EYBy153.png

    The infinitesimal Euclidean distance is ##ds^2 = dx^2 + dy^2##, but ##x^2 + y^2 = r^2## in this case, and so ##ds^2 = \frac{x^2}{y^2}dx^2 + \frac{y^2}{x^2}dy^2##. (Is this right?) It seems clear that the metric will have first, second, etc... derivatives. Now it seems that that space (surface of the sphere) has a curvature, although I have not worked on the components of the Riemann tensor. (There will be any non-zero component?)
     
  19. Feb 10, 2017 #18

    Nugatory

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    This picture makes no sense as a way to assign coordinates to the two-dimensional surface of the sphere because there are points on all three coordinate axes that do not lie on the surface of the sphere - so whatever set of points are being labeled by those coordinates, it's not the set of points that corresponds to the surface of the sphere. Furthermore, you have three coordinates not two - taking ##z=0## reduces the number of coordinates by one, but the surface that is being described by the two remaining coordinates is not the surface of the sphere, it's a plane that intersects the sphere.

    One way or another you need two coordinates, not three, that cover the surface of the sphere without picking up any points outside the sphere (informally, this means that the coordinate axes lie on the surface of the sphere). Latitude and longitude would work, as would stereographic coordinates. You can calculate the value of the components of the metric tensor for the surface of the sphere in your chosen coordinate system by using the concept that @Orodruin mentioned above: it's an induced metric once you've decide to think of the two-dimensional sphere as something embedded in Euclidean three-dimension space.
     
  20. Feb 10, 2017 #19
    Ok
    How would this induced metric look like in terms of Cartesian coordinates?
     
  21. Feb 10, 2017 #20

    Nugatory

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    You mean the Cartesian coordinates of the three-dimensional Euclidean space in which the sphere is embedded?
    Suppose you go with latitude ##\phi## and longitude ##\theta## as your coordinates on the surface of the sphere. We have (although wise people will check my algebra):
    ##x=R\cos\phi\sin\theta##
    ##y=R\cos\phi\cos\theta##
    ##z=R\sin\phi##
    relating the ##\theta## and ##\phi## coordinates of points on the two-dimensional surface of the sphere to the x,y, and z coordinates of points in the three-dimensional Euclidean space in which the sphere is embedded. Some algebra and the formula in the wikipedia article I linked will see you home from there.
     
    Last edited: Feb 10, 2017
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