## H3 A Quantitative Consideration

How far does an apple fall in one second from initial rest? Assuming for g the rounded value of 10, we get from Δh = 0.5 • g • Δt

^{2}a distance of 5 m. But one second of time on our scale diagram corresponds to a length of 300,000,000 m! If a second on the diagram has a length of 3 cm, a distance of 5 m in the diagram has a length 0.5 nm (nanometers) - which corresponds to just a few atomic diameters!

Let’s calculate the length of the radius of the time axis (or, the inverse value, its curvature) when the space-time diagram is adjusted to the strength of the gravitational field at the Earth's surface:

We express all lengths in our space-time diagram in meters. The leg OP of the right triangle ZOP has, to a good approximation, the length c • 1s = 3 • 10^{8} m (the final result will show that this approximation is actually extremely precise ...). The Pythagorean Theorem now gives us:

ZP^{2} = ZO^{2} + OP^{2} ; ( ZO + 5 )^{2} ≈ ZO^{2} + ( 3•10^{8} )^{2} ; ZO^{2} + 10 • ZO + 5^{2} ≈ ZO^{2} + 9 • 10^{16}

10 • ZO ≈ 9 • 10^{16} - 25 ; ZO ≈ 9 • 10^{15} - 2.5 and thus ZO ≈ 9 • 10^{15} m

If the curvature of the time axis fits the strength of gravity on earth’s surface, then the radius ZO must be 9 • 10^{15} meters, which is almost exactly one light year! For comparison: the nearest fixed star is about 4 light years away from us and the semi-major axis of Pluto’s orbit is only 5.9 • 10^{12} m. The correct radius is therefore immense, and the correct curvature is minuscule. We are talking about a really weak gravitational field – whose curvature cannot be detected by the naked eye. But it is this tiny space-time curvature on earth’s surface that causes freely moving objects to fall and that makes climbing stairs so strenuous!

Adam Trepczynski has created a Shockwave animation, which allows one to play with Epstein space-time diagrams - with and without gravity.