6
$\begingroup$

Imagine a room where time flows infinitely faster inside of it than outside. If you are in the room, time outside the room is stopped, relative to you.

But this room is not airtight. Air molecules inside a room are constantly in motion at velocities in the hundreds of meters per second. From my -- rudimentary -- understanding of particle physics, a problem emerges. When air molecules inside the room exit it, even for a fraction of a second, they slow to a halt due to the time distortion. Pressure can't force them back into the room, because time has stopped outside of the room. That means that air molecules can flow out, but they can't flow back in. I believe that over time, this would cause the room to depressurize to a near vacuum.

Is this an accurate assumption? How quickly would this happen, from the point of view of someone inside the room?

$\endgroup$
3
  • 1
    $\begingroup$ You may have to clarify what happens at the boundary. Is it a hard boundary? If so, you'll need to defire what happens when part of a molecule hits it but the rest of it is still in the go-time space. If it's a gradient, it may be easier to paint a clearer picture of what happens... $\endgroup$
    – Qami
    Jul 7, 2021 at 22:07
  • $\begingroup$ See my answer (and others) to Maximum Survivable Time Distortion Factor?. Note that you likely have other problems that come long before depressurization. "Heat" is the word we use when we experience a substance vibrating on an atomic level fast enough - and you just accelerated it to an infinite speed relative to outside observers. What keeps your room from exploding with enough force to crack a planet? What are the rules of your world governing the transition between the two times (which is the reason I pointed out that other Q)? $\endgroup$
    – JBH
    Jul 7, 2021 at 22:48
  • $\begingroup$ LSerni's answer is nice. But your question suggests that time passes faster inside the room, not slower outside the room (ignoring the term 'infinitely' for a moment). If this is the case then for an observer outside the room, gas molecules from inside would be impinging on tthe boundary at vastly up-scaled velocities. If their kinetic energy was dumped at the boundary then this would either turn into a lot of heat energy (speed-up ~100), intense radiation (speed-up ~10,000), or black-hole inducing relatavistic overload (speed-up > 100,000). So standing outside the room might not be too safe. $\endgroup$
    – Penguino
    Jul 7, 2021 at 22:54

1 Answer 1

6
$\begingroup$

That means that air molecules can flow out, but they can't flow back in. I believe that over time, this would cause the room to depressurize to a near vacuum.

Is this an accurate assumption?

I don't think so, not for your average room, but depends on the interface characteristics.

This is my take on your scenario...

As soon as a molecule exits the "time discontinuity" volume, its speed from the point of view of insiders drops to zero. The molecule is frozen in space, with a virtual temperature of absolute zero (no thermal agitation), and remains locked at its point of entry.

Now, this would mean that the whole room is virtually exposed to hard vacuum. And so it is - at the beginning.

But at the same time, let's imagine that in the room there is a compressed air tank that keeps pressure constant. Can the air continue flowing outwards? Forever? Obviously not, because it would quickly fill a very thin volume of time-frozen air on the room boundary, and this volume can't contain an infinite quantity of air.

Time-freezing also blocks repulsive forces, so the air would be reduced to its covolume. Since liquid air has a density of 0.87, I think we can go with a time-frozen density of about that of water.

Once the interface is saturated with time-frozen air, further molecules of air cannot penetrate the interface and bounce back inside the room.

Now the crucial element is the thickness of the interface - how far can a molecule travel before being frozen. It cannot be thinner than the width of one molecule, of course, or it would act like an impenetrable wall (no air loss) from the start.

If it is, say, one thousandth of a millimeter thick, then each square meter of interface (one million square millimeters) will eventually absorb one liter of air (one million by one thousandth cubic millimeters), and become a sheet of impenetrable time-frozen air. After that, the room will contain one liter of air less than before.

A 5 x 4 x 3 = 60 m^3 room will lose (5x4 + (5+4)*3)*2 = 94 liters of air through roof, floor and walls, which is negligible. The air would travel at high speed, supersonic but not much faster, 464 m/s; the effect would be that of a loud "thwock!", and no more.

Of course, if the interface is one millimeter deep, then it can absorb one thousand times more air before saturating, which means 94 cubic meters - and the room only holds 60. You would get an explosive decompression (the air loss should be a logarithmic curve, but it would be so steep as to make little difference).

$\endgroup$
5
  • $\begingroup$ Would it be reasonable to assume that the thickness of the interface is one Planck length? $\endgroup$
    – Tom
    Jul 7, 2021 at 22:53
  • $\begingroup$ @Tom Well, we can assume whatever we want, I suppose :-D. Functionally, the minimum thickness will be one oxygen or nitrogen molecule's worth - as soon as one molecule stops, it becomes the interface. $\endgroup$
    – LSerni
    Jul 7, 2021 at 23:10
  • 1
    $\begingroup$ "the air would be reduced to its covolume" - that's a fair assumption, but only one of the possibilities. Other options include electron-degenerate matter, neutron-degenerate matter etc... $\endgroup$
    – Alexander
    Jul 7, 2021 at 23:11
  • $\begingroup$ @Alexander I have considered those possibilities, but from the external air's point of view, the inbound molecule has only a very high (relativistic) speed. I don't think we get the kind of pressures necessary to achieve electron degeneracy between two "frozen" molecules, unless the time effect is really weird (which I'll admit is possible) $\endgroup$
    – LSerni
    Jul 7, 2021 at 23:18
  • 1
    $\begingroup$ i think your model assumes an open volume, not a slowly leaking room. You won't get explosive decompression if the room is reasonably air tight. You only loose air in those small patches which are leaking, not through the concrete or glass. $\endgroup$
    – ths
    Jul 8, 2021 at 16:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .