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

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 $m.s^{-2}$)

• The maths is on wikipedia, but for survivable gravity it's barely detectable by anything that isn't an atomic clock. – pjc50 Mar 14 '15 at 18:43
• The difference is not a function of the gravitational acceleration, but of the gravitational potential. So without knowing the gravitational potential of the room (i.e. how much energy you need to leave the room), the question has no answer (all you could do from knowing the acceleration is to calculate the difference between the floor and the ceiling of the room). – celtschk Mar 14 '15 at 19:00

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.

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.

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.