By how much exactly would the time flow be slown down inside a gravity room?

Question

A super alien warrior goes training inside a gravity room. It's a room with a flat floor, but the surface gravity of the room can be modified at the push of a button.

But under general relativity a higher gravity leads to a slower time flow.

So my question is: what is the expression of $\dfrac{t_{in}}{t_{out}}$ in terms of $g$?

$t_{in}$ is the time elapsing inside the gravity room

$t_{out}$ is the time elapsing outside the gravity room (with no gravity to influence the flow of time)

$g$ is the surface gravity inside the gravity room (in $m.s^{-2}$)

Answer 1

The easiest way to accomplish this is to have the warrior in a large centrifuge that speeds up to simulate enhanced gravity on the axis perpendicular to the spin - think of the gravitron ride - you can turn sideways on the walls at a sufficient speed.

Since your survivable g-force is limited by physiology, your rotational speed could only reach a tiny fraction of c before any living organism would be crushed into jelly striated by chemical weight.

As mentioned in the comments, there would be no appreciable time dilation when training at any survivable g force. It might be measurable in picoseconds, but nothing the average person would notice.

Addendum:

From the wiki article on gravitational time dilation:

On the other hand, when g is nearly constant and gh is much smaller than c^2, the linear "weak field" approximation T_d = 1 + gh/c^2 may also be used.

Conversely, we may approximate the effects of increased gravity as reducing the distance between the warrior and the planet (eg: everyone outside the room is at a "farther above the gravitic center of mass on the planet.")

Since gravity increases/decreases by the square/root of the distance, so a person twice as close would experience 2^2 as much gravity. Extrapolating for the desired gravity multiplier x = h^2), we have: $$T_d = 1+\frac{g*\sqrt{x}}{c^2}$$ Assuming an earth-like original gravity, moving to a 9G environment (considered the maximum survivable vertical G by modern airforce pilots), then x=9, root x=3, and the equation neatly wraps up as: $$T_d = 1+\frac{3g}{c^2}$$ $$T_d = 1+\frac{3(9.80665 m/sec^2)}{(299,762,458 m/sec)^2} = 1+3.274e^{-16}$$ For scale: $$3.156e^{-16} nanoseconds = 1 year$$ Basically, the outside world would gain ten nanoseconds a year compared to the area inside the room.

Further addendum:

If we assume that native gravity for the aliens is 1 billion times earth gravity (because they live on the surface of a neutron star?) and we're talking about a 9-fold increase in native gravity (don't forget, whatever natural physiology that the aliens have is adapted to their evolved environment), then you're still only looking at: $$T_d = 1+\frac{3g}{c^2} = 1+\frac{3(9.80665*1,000,000,000 m/sec^2)}{(299,762,458 m/sec)^2} = 1+3.274e^{-7}$$ $$3.15569e^7 seconds = 1 year$$ Which still is only a 10 second gain per earth-year for the outside world. It's going to be noticeable at that point, but you only really experience significant time dilation at points where g*h approaches C, and at that point you need to use a different equation and most large solid chunks will rip themselves apart.