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The question is somewhat related to What is the sum of the angles of a triangle on Earth orbit? but still not quite what I think of.

Consider a real world triangle made of straight steel rods which are connected and fixed with screws at the corners. When such a triangle is constructed far away from any mass, i.e. in a euclidean geometry of space-time, the sum of angles will be 180 degree.

Now bring this triangle in the vicinity of a massive object. The geometry will become non-euclidean, the sum of angles will not be 180 degrees anymore. How exactly will this effect present itself on the triangle? I can think of several possibilities:

  1. The sum of angles stays at 180 degrees but the rods will bend and/or experience tension, or the rods will not bend and only experience tension.
  2. The sum of angle deviates from 180 degree but the rods will still look straight, or the rods will bend, with or without tension.
  3. There is no visible effect because the angle measuring device is bend exactly the same way as the triangle, with or without tension in the rods.
  4. Any combination of angles, bending, tension, ...

What happens when you loosen the screws?

Please note, that this just a warm up question. What I'm really interested in is the ratio of circumference and area of a circle, which happens not to be Pi in a non-euclidean geometry. How does the effect present itself when the circle is laid with tiny tiny tiles. These tiles can be used to approximate the radius and the area, so at least one of these numbers must change when the curvature becomes effective, don't they?


Edited: Clarification on the tiles. Consider a circle on the ground that was constructed in the usual way with a peg at the center, a string, and a pen to draw the circumference. When I lay tiles (that all have the same square shape and area) on the ground and count the number of tiles that are inside the circle or touch the border, and additionally count the tiles that make up the radius, I can get an approximation of $A/r^2$ which will be $\pi$ in a euclidian geometry. Thus, when I bring the circle together with the tiles into an non-euclidian geometry, something must happen to the tiles or the circle to compensate the change in the ratio $A/r^2$. I assume that the tiles will start spreading a bit and gaps will appear between them. Right?

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I’m not sure if I understand you question good enough, so let me try.

There are several ways to look at the curvature in the vicinty of a central mass. As you talk about a triangle consider the bending of light in relation to the center of the mass. So heuristically it’s clear that the sum of the internal angles of a triangle made of intersecting null geodesics exceeds 180 degrees.

The rods of your steel triangle would feel some bending and would eventually break depending on the “strenght” of the bending. If you mean “tiny tiles” in the sense of freely falling “test particles” then again the sum of the angles of a triangle formed by these would exceed 180 degrees if released from the steel rods.

The circumference you are questioning is $2\pi{r}$, where $r$ is the coordinate radius. The relation between radial proper distance (the distance measured by rulers) and the r-coordinate follows from the Schwarzschild metric as $d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$, with $R_S$ the Schwarzschild radius. By integrating one obtains a radial distance measured by rulers expressed by the corresponding difference of the r-coordinates. The above formula shows that the radial proper distance exceeds the coordinate difference and hence the circumference measured by rulers exceeds $2\pi{r}$ as required.

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  • $\begingroup$ If I understand it correctly, the angles of the triangle will stay at a sum of 180 degrees, but the rods will bend when the curvature becomes effective. When I loosen the screws at the corners, the angles „jump“ to more or less than 180 Degrees and the rods are straight again? $\endgroup$ – Hartmut Braun Jul 28 '18 at 7:18
  • $\begingroup$ Rods which are falling freely in horizontal position will feel compression due to tidal gravity. $\endgroup$ – timm Jul 28 '18 at 11:27

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