Can we see the curvature of a surface?

After reading the Feynman lectures' (chapter 42, Vol.2) , it had me thinking if it is by any way possible to measure the curvature of a surface (think, surface of earth) just by observing the nature of the horizon from, for instance, a beach?

How would the nature of the horizon differ, if one were observing from a beach on a flat surface (for instance, assume a hypothetical flat earth) ?

I would like to add that I am a total stranger to advanced geometry, non-Euclidean spaces and the like.

Image credits: Feynman Lectures Vol.2

• You'd see yourself as if at the bottom of a round dish. Commented Nov 2, 2019 at 5:06

I tried a geometric approach to answer your question. Assuming your planet doesn't have any atmosphere for refracting light and that the planet is perfectly spherical, it is possible to calculate the radius of the planet using the angle of depression of the horizon from the tangent of the planet drawn from the point where you stand. If you could somehow find out this angle, this is how to find out the radius of the planet.

The red line represents You.

$$h$$ :Your height

$$Blue-line$$: Tangent drawn on the sphere at the point where you stand

$$Dark-blue-line$$: Tangent drawn at the point you see as a horizon.

Note that the two blue lines are equal in length by geometry. From the $$\Delta$$ AOB, you get: $$(R+h)sin\theta=R$$ $$R=\frac{hsin\theta}{1-sin\theta}$$ Putting the values of $$h$$ and $$\theta$$, you can get the radius of the planet.

For a spherical planet, the horizon would go lower as your height increases.

Putting the radius of curvature equal to infinity in the equation ,$$\theta$$ turns out to be $$90$$ degrees irrespective of the value of $$h$$.

So for a very large flat planet, however high you go, you will still see the horizon near your eye level (As described in one of the comments above, you would feel as if you are in a very large round dish)

• How about when the radius gets extremely large (not infinite though)?
– Maan
Commented Nov 2, 2019 at 12:00
• I would go with (R + h) sin Ɵ = R Commented Nov 2, 2019 at 17:52
• In that case, I'd continue with the same general idea, except that, if it was a sphere (and even if the sphere was not the surface of a ball), the inward focusing of some of the gravitational effects of whatever particulate composition it would have would cause the limitations I'd described to apply earlier, unless perhaps the gravitation would've attracted enough energetic particles for their refractions of each other to reverse those effects, as in Eddington's 1919 observational proof of GR. Commented Nov 2, 2019 at 20:30
• @Edouard , I'd say you should give a shot at answering the question, your suggestions are interesting
– Maan
Commented Nov 3, 2019 at 3:54
• @R.w. bird Yes, thanks for pointing it out. Commented Nov 3, 2019 at 5:23

No. Spacetime is locally flat (Minkowskian). Same with the surface of a sphere. In order to detect curvature, you either have to move out of the manifold; which you can do by climbing a mountain on Earth, but you can't do in spacetime; or you need to take a survey. Read the introductory chapters of Spacetime Physics by Wheeler and Taylor, and, more importantly, of MTW's Gravitation.

But you ask a good question. The negative answer is the essence of Einstein's Equivalence Principle.

In response to comments, let me add that: by survey I really mean what those guys with site glasses and tripods do. It was how Gauss came up with his method of intrinsically determining curvature. The book itself is quite advanced, but the introduction explains how this relates to the equivalence principle without too much math, so I suggest reading the first chapter of Gravitation and Inertia The math in the second chapter gets a bit more intense, but the verbal part of the discussion of the equivalence principle stands pretty much on its own. The seminal thought experiment is Einstein's elevator. If you were in deep space, confined to a windowless elevator, you would not be able to do a local experiment to prove that you were not freely falling in a "gravitational field". Just as, if you are standing on the shore line, you will not be able to see the curvature of Earth. (within certain limits of precision).

The larger the elevator, the more likely you will be able to detect tidal forces (curvature,) because you can measure the relative acceleration of more distantly separated test objects.

Imagine you were on a perfectly smooth planetary surface, and you wanted to paint rectangular parking spaces. If you restrict your parking lot to a few hundred meters, you will get away with using a square and some string to map out the lot. If you want to do the same over several miles, you will eventually run into problems because lines which are parallel at on location will become closer at other locations.

• Maybe I should ask this as a separate question, but, perhaps in trying to grasp how the concept of "locality" could be grasped within a spacetime (GR's) which is regarded as infinitely expansive to the future (and felt by many to be symmetrically infinite to the past as well, in explanations as to how the observed gravitational collapse of stars might fit into the panorama), I've had a problem with understanding how a radial phenomenon like gravity could be compatible with a potentially non-radial one like acceleration alone. (I had, and lost,1 good explanation of that.) Commented Nov 3, 2019 at 15:17
• Oh, I think I just got your point: The survey would provide an estimate of whatever proportion of our Observable Region is occupied by material bodies (stars and planets), and any smaller proportion (like Earth's volume) would be "local", right? Commented Nov 3, 2019 at 16:02
• Enough time spent wearing out keyboards everywhere! Since you've answered both the OP's question & mine about the Equivalence Principle, you've got my vote! Just let me know, if it was cast for the wrong reasons. Commented Nov 3, 2019 at 16:17

When one lives next to the sea, on a calm day, the curvature is evident if a sails ship goes away from you. The ship and then slowly the sail disappear.

You could approximately measure the curvature assuming a spherical shape, by taking a calm day when the sea is almost like a mirror, a sailboat with a motor going at fixed velocity, and timing when the sail disappears.

Or being convinced by the disappearance of ships at a distance that there is a curvature , use Eratosthenis method ( 200BC) to measure the circumference of the earth.

Edit: While driving east during sunset on a dry clear day in Athens Greece, I saw the shadow of the earth, which was an arc, similar to this:

Given the correct distances , one could calculate the radius of the sphere making the shadow using the arc.