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From the Wikipedia link for Geometry:

Geometry (Ancient Greek: γεωμετρία; geo- "earth", -metron "measurement") is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space.

Up to the work of Riemann and Gauss, this definition would have made clear to me examples of geometrical objects: a line, square, cube, hypercube; each of these possessing geometrical properties such as the number of sides, angles between faces, the dimension of space that contains them etc. Hence a geometrical object was a set of measurements associated with an object using a ruler.

After the work of Einstein and Minkowski who showed that time and space were a part of one another, would it be correct to say:

  1. a geometrical object is a set of measurements associated with an object of distance and time using a ruler and a clock ?

  2. Geometrical objects includes an interval of time, the invariant space-time interval?

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You're mixing maths with physics, which in this case is not possible. – jinawee Jun 22 '13 at 22:39
@jinawee the physical world has a geometry that can be modelled using maths, making it possible to mix maths with physics. – Larry Harson Jun 22 '13 at 23:02
Suppose that Planck's length would be the minumum distance in the Universe. A mathematician would still talk about points, triangles and all sort of stuff that doesn't exist. You can talk about the geometry of the Universe, but that won't modify the classical geometry in maths . – jinawee Jun 22 '13 at 23:14
@jinawee A physicist can still approximate these objects on scales where the error becomes smaller. Compared to the size of the earth, a coin is a good approximation as a point when calculating the distance between London and New York using geometry. – Larry Harson Jun 23 '13 at 0:45

In SR, you can have geometrical figures in spacetime. You don't need both rulers and clocks to define them. All you need is a clock. There is a nice presentation of this kind of thing in Laurent 1994.

In GR, you have a spacetime whose curvature varies from point to point. In this kind of spacetime, you can't in general transport geometrical figures from one place to another (and you also can't scale them). There is no notion of congruence. For these reasons there isn't much interest in studying geometrical figures for their own sake.

The ancient Greeks conceived of geometry in terms of figures and congruence, but today the notion is much more general than that. For instance, you can have projective geometry, which has no system of measurement, or geometries with finite numbers of points.

Bertel Laurent, Introduction to spacetime: a first course on relativity.

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Thanks for the reference, I'll take a look. – Larry Harson Jun 23 '13 at 13:39

In relativity, yes you are on to something in that time is pretty much just another dimension like space, and clocks are just a special type of ruler.

However, a geometrical object is a bit more than what is stated. Physicists use the term loosely for a collection of measurements that, when taken together and interpreted correctly, is in some sense invariant. That is, the numbers themselves might change, but their collective interpretation does not, when you change coordinates.

The archetypical example is a vector/"arrow" in $\mathbb{R}^2$. Say we have a vector $\vec{v}$ representing a change of $+1$ in $x$ and $+2$ in $y$. We might write1 $$ \vec{v} \stackrel{(x,y)}{\longrightarrow} (1, 2). $$ However, we could also parametrize the plane with another coordinates, say $(w, z)$, related to the original coordinates by $w = 2x$, $z = -y/2$. In this basis we have $$ \vec{v} \stackrel{(w,z)}{\longrightarrow} (2, -1). $$ The numbers have changed, but each set of numbers, when interpreted with the proper basis in mind, refers to the same geometrical object, $\vec{v}$.

1 What makes this confusing is most authors do not explicitly say which basis they are in when writing formulas like this one. In fact, they often use "$=$" rather than "$\to$", and so you'll see confusing things like $\vec{v} = (1, 2)$ and then three lines later $\vec{v} = (2, -1)$. Clearly as mere tuples $(1, 2) \neq (2, -1)$. They are only equal as geometrical objects when each is ascribed the appropriate interpretation.

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