1. Nov 30, 2011

### mrspeedybob

Suppose I construct a metal triangle in flat space the sum of the interior angles will be 180°. If I then move the structure into a curved space in which the sum of the interior angles of the triangle will be greater then 180° do the corners of my triangle experience stress? Or, do I simply measure a wider angle at each corner in the curved space because that is the nature of the space. Thinking this through I have 2 conflicting lines of reasoning...

1. If the corners of the structure experience stress they will flex. Now I have a force acting through a distance so work has been done to the structure. Where did it come from? If the triangle and I start out in flat space and I push it toward a region of curved space then the work must come either from the KE that I put into it when I pushed it or from the stress-energy that is curving the space. The first possibility would violate conservation of momentum. The second I can't wrap my head around well enough to determine if it's plausible yet.

2. If I simply measure a wider angle at each corner because that is the nature of the curved space, how do I measure it? If I take a protractor into the curved space along with my triangle wouldn't the same thing happen to my protractor that happens to my triangle, thus the correlation between a corner of triangle and a mark on the protractor would not change.

2. Nov 30, 2011

### PAllen

You can answer this question qualitatively (and to a very good approximation for weak fields) with Newtonian theory. GR simply provides a different conceptual model (of course, also different predictions for strong gravity).

Thus, the triangle would feel stresses whether held stationary or in free fall (assuming it is big enough to experience the gradient in potential). In weak gravity, if it is 'reasonably rigid', you would not measure angular change - you would have stress instead.

If it was less rigid, so it deformed, conservation of energy and momentum would be preserved taking into account the source of gravity. In GR, you would have exact conservation only if you take gravitational radiation into account, and if the spacetime is asymptotically flat; otherwise these concepts cannot be adequately defined in GR.

Last edited: Dec 1, 2011
3. Dec 1, 2011

### pervect

Staff Emeritus
It's easiest I think to envision the problem in flat space-time, where you have a flat triangle on a a plane, and then you draw the same triangle on a sphere.

I'm not sure if this is what you're envisioning, but it seems relevant and easier to answer.

I would assume that you are "moving" the triangle in such a way that the lengths of its sides remain constant. So you can take triangle, and project it onto the sphere and keep all three sides the same length and have a triangle with the same side lengths on a sphere.

You don't really need to worry about "stress", but you probably want to have some concept of preserving the lengths of the sides. If you think of the triangle as being made out of wire that can bend, but can't change length, that's close to ideal - the wire will have to bend some to go from a plane to a sphere. Of course because the wire can bend you also need to make sure that the wire marks out a great circle (the geodesic equivalent of a straight line) on the sphere, unless you somehow imagine a wire that can bend in one direction but not another (possible, I suppose, but I can't quite imagine how to make such a thing).

As far as measuring angles, on a plane you know that if you draw an angle theta, the arc length of a circle of radius r inside the angle will be r * dtheta , if theta is measured in radians.

On the sphere, if you use a large circle, the arc length will be r * (circumference / 2 pi r) * dtheta instead, which you can see approaches the plane value if you use a small circle, and needs correction if you use a big circle.

You can even work out the exact value if you like, by picturing a plane slicing your sphere, and looking at the ratio of the subtended arc (which will be r on the sphere) to the chord, since the circumference of the circle will be 2*pi*length_of_chord.