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Is arithmetic still true in a gravitational field

  1. Dec 24, 2005 #1
    It is well-known that Euclidean geometry does not hold in a gravitational field.

    Because of equivalence, an entity in circular motion is effectively the same as an entity in a gravitational field. Now because length shortens in the direction of motion in accordance with special relativity, the circumference of a rotating circle will shorten according to someone stationary relative to the rotating circle. But because the diameter of this circle is not moving parallel with the direction of motion, the diameter stays the same. Thus the relation c = d * (pi) no longer holds. Theoretically, if such a circle is rotating at the speed of light, the diameter will be infinitely bigger than the circumference. This is why Euclidean geometry does not hold in a gravitational field. But since there is at least some gravity anywhere in the universe, Euclidean geometry does not hold anywhere in the universe.

    Now geometry is merely a spatial way of expressing arithmetical relations. Conversely, arithmetic is merely a Platonic way of expressing geometrical relations. Consequently, it is my opinion that Platonic arithmetical relations will also not hold in a gravitational field. Allow me to demonstrate:

    Take an equilateral triangle and give it a constant circular rotation in the plane of its sides and centered on its center of "gravity" (don't take that metaphor too seriously though). Let's say the length of each side (before the rotation) is 2 units. Now draw a straight line down the middle of the triangle such that it divides the triangle into two right triangles. Because this line is not travelling parallel in the direction of motion, its length will not shorten relative to someone standing stationary outside the triangle. Let's also say that the rotation is sufficiently fast enough to cause the triangle's sides to shrink by a factor of two relative to someone stationary. The length of the "dividing" line is the square root of 3. This is because of Pythagoras' Theorem such that (square root of 3)^2 + 1^2 = 2^2 (before the rotation). Thus square root of 3 = square root of (2^2 - 1^2). But after the rotation, we would get this relation: square root of 3 = square root of (1^2 - .5^2), or square root of 3 = square root of .75. Consequently, because Pythagoras' Theorem is an arithmetical relation as well as a geometrical one, arithmetic must not hold in a gravitational field. One might argue that the angles of the triangle changed during the rotation, but since all sides shrink by the same factor, it is still an equilateral triangle and the dividing line also still creates two identical right triangles.

    But since there is at least some gravitational field everywhere in the universe, then arithmetic must not hold anywhere in the universe. If I am wrong, will someone please correct me?
     
  2. jcsd
  3. Dec 24, 2005 #2

    selfAdjoint

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    You are trying to apply the ususal Pythagoras rule to get your square root formulas. It is this rule that fails, not arithmetic.
     
  4. Dec 26, 2005 #3

    HallsofIvy

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    Arithmetic has nothing to do with physics. It doesn't matter what the gravitational field is the laws of arithmetic, which are, after all, simply generally agreed upon definitions, are still true.

    The "Euclidean geometry" model for spacial geometry does not hold in the presence of a gravitational field, as selfAdjoint said, so the Pythagorean theorem does not hold.
     
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