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Varying Gravitational Field - Invariant Tetrahedron?

  1. Nov 8, 2011 #1
    Varying Gravitational Field - Invariant Tetrahedron??

    Classical Theory of Fields, Landau Lifgarbagez, page 246:

    "Strictly speaking, the number of particles should be greater than four. Since we can construct a tetrahedron from any six line segments, we can always, by a suitable definition of the reference system, make a system of four particles form an invariant tetrahedron. A fortiori, we can fix the positions relative to one another in systems of two or three particles."

    This is a footnote to a section where the change in reltive spatial distances between bodies in a varying gravitional field is discussed. I have no idea what it means by it. Could someone explain? I have never been so puzzled!!
  2. jcsd
  3. Nov 8, 2011 #2


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    Re: Varying Gravitational Field - Invariant Tetrahedron??

    First, analogy. On a 2-sphere (surface of a ball), there is no intrinsic way to distinguish that geodesic segment curves. Intrinsically, it is indistinguishable from a Euclidean straight line. Further, make two geodesic segments intersect at right angles. There is still no 'internal to 2-sphere' way to distinguish this from Euclidean. Add a third geodesic segment closing a triangle - now you can distinguish. The angles don't sum to 180, and the lengths don't match Pythagorean theorem.

    In a semi-riemanning 4-manifold, it turns out (if you are free to choose your spacelike slices -i.e. coordinates) you can always have tetrahedron that is indistinguishable from Euclidean. Curvature of space-time does not force you to have a non-Euclidean tetrahedron. You need 5 points before you are forced to have some Euclidean violation.

    Since your book leaves this to a footnote, see if you can find (in a library):

    "Relativity, The General Theory" by J. L. Synge

    He provides a detailed derivation of this fact.

    I don't know of any readily available source with this derivation (e.g. MTW doesn't mention it).
  4. Nov 8, 2011 #3


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    Re: Varying Gravitational Field - Invariant Tetrahedron??

    Sort of a sidenote (I'll have to admit to not being able to obtain Synge to read it either, though it sounds very interesting).

    If you don't measure the angles, and only measure the sides, you'll need four points to detect the intrinsic curvature of the sphere's surface.

    If you draw a Euclidean triangle on a plane with three sides a,b, and c (which must satisfy the triangle inequality), you can construct a triangle with sides a,b,c on the sphere with the same lengths (we're not measuring the angles for this exercise).

    You start off with drawing a line segment that has the length of the 'a' side, then draw one circle of radius b centered at one end of the line segment you just drew, then draw another circle of radius c centered at the other end. The two circles will intersect at a pair of points (usually, only one point if the triangle is degnerate), giving you the desired triangle.

    By only measuring lengths, you can still tell that the sphere is curved by drawing a square, which for the purposes of this exercise is defined as a quadrilateral with four equal sides and two equal diagonals.

    When you draw the square, you'll find that on the sphere the ratio of the diagonal to the side is not sqrt(2).

    If you take the sum of the squares of the diagonals, and subtract the sum of the squares of the sides, you'll get zero on a flat plane, and a non-zero number on a sphere, i.e if the side is s, the diagonal on a plane is sqrt(2) s, so the sum of the squares of the two diagonals is 4 s^2, the sum of the squares of the four sides (s^2+s^2+s^2+s^2).
    Last edited: Nov 8, 2011
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