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Euclidean Geometry

  1. Oct 14, 2009 #1
    Considering that space is not curved or warped (as some pop books will falsely lead you to believe) why is that Euclidean Geometry is not true in the real world?

    I mean light bends in space because it falls in a gravitational field like everything else (because it has energy which is equivalent to mass), why choose them to draw straight lines when we know they wont travel in a straight line if they have a huge mass near them, if i imagine straight lines and a triangle in space i dont see any reason to think that the sum of their internal angles will not be 180.
    i mean i dont make triangles on earth by throwing baseballs in the sky.
  2. jcsd
  3. Oct 14, 2009 #2

    Jonathan Scott

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    The curvature of time and space is very small. Consider the gravitational field g of the Earth, about 10ms-2. The corresponding radius of curvature R of space is given by R = c2/g, which is about 9 x 1015m (I think that's about a light year).

    Since we effectively move through time at c, this is why the resulting accelerating for a body at rest is c2/R which gives g.

    Even the curvature of space close to the sun is difficult to observe; Eddington's famous expedition managed to observe a very slight shift of a star (less than a thousandth of a degree) near the sun during an eclipse in 1919, confirming the predictions of Einstein's General Theory of Relativity.
  4. Oct 14, 2009 #3


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    I find this difficult to understand. Whether space is curved or not depends upon what geodsics you have. Now, having stated that space is NOT curved, you are asking why someone (who?) draws the path of light in straight lines.

    You appear to be saying that the "curvature" of space is an artifact of trying to use the path of light as a geodesic rather than straight lines.

    I guess my question then is- how do YOU define straight lines in space? What properties would they have that would allow us to identify straight lines?
  5. Oct 14, 2009 #4


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    It's not "pop books" that say that, the basic idea of general relativity is that mass and energy curve spacetime (giving it a different metric than flat spacetime, with the metric being used to measure the 'length' along any path through spacetime, which in the case of timelike worldlines corresponds to the proper time along that worldline), and that objects which aren't under the influence of non-gravitational forces will follow geodesic paths in curved spacetime (for timelike worldlines, these are paths that locally maximize the proper time--any 'nearby' worldline between the same two points will have a smaller value for proper time along it than the geodesic worldline). Spatial curvature is a bit less objective since you can foliate 4D spacetime into a stack of 3D spacelike surfaces in a variety of different ways (corresponding to different definitions of simultaneity), but once you have a foliation, if you take spacelike paths that lie within a given spacelike surface, you can use the metric to measure the length along them too, and the metric will usually be different than the metric you'd see in 3D Euclidean space (so there'd be no way to map points in the spacelike surface to points in a 3D Euclidean space such that the length of every path in the spacelike surface would match up to the length of the corresponding path in Euclidean space). In differential geometry terms, this means that the geometry of the spacelike surface really is different from that of Euclidean space.
    Last edited: Oct 14, 2009
  6. Oct 14, 2009 #5


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    Space is curved in the same sense a ball is spherical. Curvature and spheres are both just mathematical concepts, but they describe real things quite well.
    Euclidean Geometry is just another mathematical concept which describes some real things quite well.

    As I said in the other thread : you don't get the correct trajectory of an moving object from that local fall acceleration alone. The overall observed trajectory (not only of light) is also affected by the spatial curvature:

    You confuse things here. The baseballs take the shortest paths in curved space-time not in space.

    If you draw triangles on earths surface, by taking the shortest paths within the surface between the points, you will not get 180 as the sum of angles. Curved 3D space in not different: you connect 3 points by the shortest paths within the space, and get a sum of angles different from 180.
    Last edited: Oct 14, 2009
  7. Oct 15, 2009 #6


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    Ah, it is starting to make sense. Previously, xMonte asked if space were really curved and was told that that was a mathematical model.

    He seems to have concluded from that that space is NOT curved. I'm afraid the problem is that xMonte just doesn't understand the roll of "mathematical models" in physics. The fact that something is a mathematical model does NOT mean that what it models is not real!
  8. Oct 15, 2009 #7
    Yep you are spot on and thanks for clearing things up, so around a huge mass space "is" curved and there is a mathematical model for it (GR).
    But in my defense i asked specifically

    Is space really curved/warped in presence of mass or is it just a metaphore, is it more than a mathematical concept?

    and i got the answer
    It is a mathematical model.

    which led me to believe its not really curved just mathematical interpretation.
  9. Oct 15, 2009 #8
    There are theories of gravity in a plane, Minkowski space-time (RTG) where the light is just affected with the gravitational force. Having constant and plane geometry is much better than working in varying, different geometries.
  10. Oct 15, 2009 #9


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    Imagine you didn't have eyes, but were capable of using measuring sticks to measure various distances on the surface of the earth. Then eventually, you will have a theory where you would imagine that you were on the surface of a sphere. This would be your 'mathematical model'. You could say that the surface isn't 'really curved', just 'mathematical interpretation', but that would be silly. It is the same in general relativity. We can't 'look' at spacetime and 'see' its curvature, but the fact that this model fits all observations (just like in the example above) means that it is 'real'.
  11. Oct 15, 2009 #10
    i agree 100%
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