# Virtual tidal effect within an accelerating ship

## Summary:

Tidal effect inside accelerating cylinder re: equivalence not so equivalent?
Does a sort of virtual tidal effect exist inside an accelerating body (space ship) similar to the way the tidal effect is present while rest on a massive body? Or could an accurate enough 'thought experiment' measuring device figure out if it's in an accelerating ship or resting on a planet, by measuring tidal effect or no effect?

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Nugatory
Mentor
Or could an accurate enough 'thought experiment' measuring device figure out if it's in an accelerating ship or resting on a planet, by measuring tidal effect or no effect?
Yes, an accurate enough experiment will be able to distinguish between the two cases. Hold your arms outstretched with a small object in each hand, then drop the objects, let them fall to floor, measure the distance between the two landing points. In the accelerated spaceship this distance will be exactly equal to the distance between your hands (objects fall on exactly parallel paths) while on the surface of the earth that distance will be slightly less (objects fall on converging paths that intersect at the center of the earth).

This doesn’t violate the equivalence principle, which informally says that the two cases are equivalent as long as the times and distances involved are small enough. More precisely, the difference between the two cases approaches zero as the distances get smaller, so no matter how much we improve our instruments we can always construct a situation in which the cases are equivalent.

anuttarasammyak
Gold Member
Rindler coordinates, https://en.wikipedia.org/wiki/Rindler_coordinates, describes your spaceship case.
As written in that wiki, proper acceleration is ##\alpha_i## .at ##x=\frac{c^2}{\alpha_i}##. The inverse of proper acceleration is proportional to "height", so tidal effect is obvious. ##\alpha_i## is infinity at x=0, i.e. event horizon.

PeterDonis
Mentor
2019 Award
The inverse of proper acceleration is proportional to "height", so tidal effect is obvious
What you are describing is not a tidal effect. There is zero tidal gravity in Rindler coordinates since they are coordinates on flat Minkowski spacetime, which has zero Riemann curvature and therefore zero tidal gravity.

This is a good illustration of the fact that tidal gravity is not exactly the same as a difference in the "acceleration due to gravity" between two spatially separated points.

anuttarasammyak
anuttarasammyak
Gold Member
@PeterDonis Thanks. I write below to confirm no tidal force work in Rindler coordinate FR and rotating FR.

Say IFR, in which the rocket is accelerating, has particles at rest to show the lattice points. The rocket passengers observe the particle lattice swell, stop swelling and shrink in x-direction, say the direction of acceleration. Say there are bars connecting the particles. The bars swell, stops swelling and shrink. All the bars are in free motion in the rocket FR and IFRs, and get no over torsion or break.

Say IFR, in which a merry-go-round is rotating conterclockwise, has particles at rest to show the lattice points. The merry-go-round passengers observe the particles go round clockwise. Say there are bars connecting the particles. The bars keep going round clockwise. All the bars are in free motion in merry-go-round FR and IFRs and get no over torsion or break.

PeterDonis
Mentor
2019 Award
I write below to confirm no tidal force work in Rindler coordinate FR and rotating FR.
Whether or not tidal gravity is present does not depend on what reference frame you adopt. It depends on whether or not spacetime is curved. Your scenarios take place in flat spacetime, so there is no tidal gravity.

anuttarasammyak
pervect
Staff Emeritus