Can We Define Angles in Flat Spacetime Using Geodesics?

[pervect]

In another thread, it was asked if we could use the angular deficit idea to determine curvature not in space, but in space-time.

My idea to attempt to proceed along these lines would be to generalize the idea of angle, but I don't have anything that I feel I can point to.

As a starting point, I'd like to ask - if we have a flat Minkowskii spacetime, and form a triangle form one timelike geodesic and two null geodesics, is there a meaningful concept of the "angles" of this triangle that sum to 180 degrees? Possibly based on using the dot product to determine the angle?

One example of such a triange would be setting the three points of the triangle as (t,x) given by (0,0) (2,0) (1,1)

 

[robphy]

Of course, the generalization of angle between timelike-vectors is the "rapidity".Possibly interesting reading (on my to-read list):

http://projecteuclid.org/download/pdf_1/euclid.mmj/1029002964
Michigan Math. J., Volume 31, Issue 1 (1984), 77-81.
The Gauss-Bonnet theorem for $2$-dimensional spacetimes.
Graciela S. Birman and Katsumi Nomizuhttp://nyjm.albany.edu/j/2006/12-8v.pdf
New York J. Math. 12 (2006) 143–155.
The Gauss–Bonnet theorem for Cayley–Klein geometries of dimension two
Alan S. McRae

 

[pervect]

The rapidity maps -c<v<c into -$\infty$ to +-$\infty$ right? So I'm not sure if the rapidities in a spacetime triangle sum to 180. I suspect not, but I haven't tried to consider a case without troublesome infinities as in the one I mentioned.

 

[PeterDonis]

The rapidity maps -c<v<c into -$\infty$ to +-$\infty$ right?

Yes; the mapping is the $\tanh$ function (if we're mapping rapidities to velocities, or its inverse if we're mapping velocities to rapidities).

I'm not sure if the rapidities in a spacetime triangle sum to 180.

I don't see how they could, in general, since, as you note, the range of rapidities is infinite, not bounded.

I think there's also another issue, which is that just defining the angle as rapidity between timelike curves is not enough; you also need to have some notion of "supplementary" rapidities (the analog of 180 degrees minus an angle in ordinary geometry). I don't see how to define such a notion, because I don't see any natural analog of "180 degrees" in rapidity space.

For example, consider the triangle with (t, x) vertices A (0, 0), B (1, 0), and C(3, 1). All three sides of this triangle are timelike, and two of the angles are "acute"--the angles at vertices A and C (i.e., these angles are just the "rapidity angles" between the tangent vectors of the two sides that meet at that vertex).

So computing these two angles is easy (I'll use $\eta$ for the rapidities):

Angle A: $\eta_A = \tanh^{-1} \left( \frac{1}{3} \right) \approx 0.34657359027997264$

Angle C: $\eta_C = \tanh^{-1} \left( \frac{\frac{1}{2} - \frac{1}{3}}{1 - \frac{1}{6}} \right) = \tanh^{1} \left( \frac{1}{5} \right) \approx 0.2027325540540822$

But the third angle, the one at B, is "obtuse"; it is not the same as the "rapidity angle" between the tangent vectors of side AB and side BC (taking both as future-pointing), it is the "supplement" of that rapidity angle. I.e., it is the supplement of

Angle B': $\eta_{B'} = \tanh^{-1} \left( \frac{1}{2} \right) \approx 0.5493061443340548$

You will note, btw, that A + C = B', which is to be expected if you think about what the angles mean physically. But how am I supposed to get the "actual" angle B in the triangle from B'?

 

[DrGreg]

Instead of using interior angles adding up to 180°, use the equivalent form that the exterior angle equals the sum of the two opposite interior angles. (What Wikipedia calls the "high school exterior angle theorem".)

That version works for rapidities (for timelike geodesics in flat spacetime) too, in the case when all three "angles" are "acute".

But that doesn't help for null geodesics.

 

[PeterDonis]

That version works for rapidities (for timelike geodesics in flat spacetime) too, in the case when all three "angles" are "acute"

Can you have a triangle in flat spacetime with all three sides timelike and all three interior angles "acute"?

And is that condition necessary in the spacetime case? In the example I gave, only two of the interior angles are "acute"; the third, angle B, is "obtuse". But my example satisfies the theorem: as I noted at the end of my post, A + C = B', where B' is the exterior angle and A and C are the two opposite interior angles (where "angles" here means rapidities).

 

[DrGreg]

Can you have a triangle in flat spacetime with all three sides timelike and all three interior angles "acute"?

And is that condition necessary in the spacetime case? In the example I gave, only two of the interior angles are "acute"; the third, angle B, is "obtuse". But my example satisfies the theorem: as I noted at the end of my post, A + C = B', where B' is the exterior angle and A and C are the two opposite interior angles (where "angles" here means rapidities).

No, sorry, what I meant was two acute interior angles and one acute exterior angle. And specifically for the all-timelike case, to be able to calculate rapidity. All three rapidities will be between pairs of future-pointing timelike vectors.

You are right that you don't need to specify acuteness in the all-spacelike case.