Suppose I'm trapped in a box, and the box is $3\ \mathrm{m} \times 3\ \mathrm{m} \times 3\ \mathrm{m}$. I designate one vertex to be $(x,y,z)=(0,0,0)$ and the opposite one as $(x,y,z)=(3,3,3)$. I'm standing in the corner where $x=0$ and $y=0$ and feel a gravitational acceleration vector of $ \left \langle 0,0,-10 \right \rangle $.

I expect I would be able to tell the difference between acceleration and a gravitational field by measuring the acceleration vector at $x=3, y=3$. Because gravity is center-seeking, I would expect the apparent acceleration vector to be something like $\left \langle -\sqrt{18},-\sqrt{18},-8 \right \rangle$. Whereas if it's just acceleration in no gravitational field, I would expect it to stay $\left \langle 0,0,-10 \right \rangle$. The change might be really slight, depending on how far the gravitational center is, but any change would be theoretically measurable.

It seems likely that I don't understand something. Either gravity, or acceleration, or something fundamental about the equivalence principle. What am I missing?


The equivalence principle is a local statement of equivalence. The setup you designed is manifestly non-local: you're comparing the perceived acceleration at different points. (Although I think in a $3m^2$ box and any significant gravitational field, the difference in acceleration measured from one corner to the other would be infinitesimally small). This is along the lines of tidal forces not being included in the regime of the equivalence principle, as they both violate the locality requirement.

See the definitions of the various EP's, which state where they're applicable (and the locality conditions): https://en.m.wikipedia.org/wiki/Equivalence_principle

  • $\begingroup$ Thanks. I suspected it might only apply locally, but I wanted to be sure. $\endgroup$ – SarcasticSully Dec 21 '20 at 18:35

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