Question

What is the sum of all the virtues of the earth?​

Answer 1

Step-by-step explanation:

Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles.

The key fact we need to know is that the deflection angle θθ of a light ray in the Schwarzschild metric is given by:



θ=4GMc2r(1)(1)θ=4GMc2r

Strictly speaking this is an approximation that fails when the curvature is high, but at the Earth-Sun distance the curvature is so small the the approximation is essentially exact.

The reason this applies to your question can be seen by drawing out the experiment you describe:



We take three points around the Earth's orbit that form the corners of an equilateral triangle and shine the light ray between the corners. Because the light ray is deflected by the Sun we have to shine the light ray outwards a bit (I've grossly exaggerated in the diagram!). Equation (1) gives us the total deflection of a light ray that comes in from infinity and departs to infinity. Our light ray will have a deflection smaller than this because it neither comes from infinity nor ends up at infinity, but we can use equation (1) as an upper limit for the deflection angle we observe. So let's calculate that upper limit.

The angle between the light ray and the side of the triangle is just θ/2θ/2, and there are six such angles, so the sum of the interior angles exceeds ππ by 3θ3θ. The upper limit for the sum ΘΘ is therefore:

Θ≤π+24GMc2rΘ≤π+24GMc2r

where rr is the radius of the Earth's orbit and I've made the approximation that the distance of closest approach is r/2r/2. It remains only to feed in the values for the various parameters, which gives us:

Θ≤π+0.00000024