1
$\begingroup$

The Scenario:

A group of scientists on a research space station are conducting tests on a Bose-Einstein condensate. They can't figure out why it is that every time they get their condensate to form, the pressure in their vacuum container suddenly spikes as new gas is added to the pressure vessel.

It turns out that the problem is one of wormholes - the condensate is displaying negative-matter properties, and stabilising quantum-level wormholes long enough for atoms of gas on the far side to the wormhole to enter the experimental area.

I'd like the researchers to realise this when the pressure vessel explodes, and they realise it's because the other end of the wormhole had opened in a star's core, or perhaps a neutron star's crust or similar, allowing the insanely energetic and high-pressure material to flow into the test chamber and destroy everything.

Trouble is I'm not sure what material to use, or how much of it to use, and I don't know what calculations to do to figure it out.

Assume:

  1. A cylindrical pressure vessel about 1 metre x 0.5 metres, made of steel and thick glass;
  2. Ideally it should be a survivable blast for those about ten metres from the explosion;
  3. The cause of the explosion should be confirmable.

Where should the wormhole lead, and how much material should it transport?

$\endgroup$
  • 1
    $\begingroup$ @sphennings Of course it can lead wherever I want it to. The question is where should it lead to cause the explosion. This is a detail I need to build up the setting, not telling a specific story. $\endgroup$ – Werrf Jun 5 '17 at 12:59
  • 1
    $\begingroup$ Since you can arbitrarily adjust the time during which the wormhole remains open, and the speed at which matter flows through, the size of the explosion is whatever you need it to be, and the reader is unlikely to check your math on that. But the chance of a random wormhole leading to anywhere but empty space are literally astronomically low, and the chance of that wormhole then creating an explosion that is both noticeable and survivable even more so. $\endgroup$ – HugoRune Jun 5 '17 at 13:26
  • $\begingroup$ @HugoRune Upvote for a contextually proper literal use of "astronomically low." $\endgroup$ – Isaac Kotlicky Jun 5 '17 at 14:18
  • 2
    $\begingroup$ "Ideally it should be a survivable blast" Can't they just check the video to figure out why they're cleaning theoretical physicist off the walls? $\endgroup$ – Isaac Kotlicky Jun 5 '17 at 14:22
  • 2
    $\begingroup$ @IsaacKotlicky That's experimental physicists they're cleaning off the walls. Sensible theoretical physicists will be back home on Earth working on papers on the physics of what happened on the research space station. $\endgroup$ – a4android Jun 6 '17 at 1:34
3
$\begingroup$

Interesting question. I'm reasonably certain that the answer boils down to a property of matter called the equation of state.

An equation of state is a relation between several thermodynamic variables, typically pressure ($P$), density ($\rho$), and temperature ($T$). You might have heard of some simple ones, such as the ideal gas law. If we know the equation of state of a substance, we can figure out how it will behave under certain circumstances.

The matter in a normal main sequence star can be described, as a good approximation, by the ideal gas law: $$P\propto\rho T$$ where the constant depends on the composition of the fluid.

This equation of state to a nice effect informally called the "thermostat mechanism", which stabilizes the star against instabilities. If the temperature rises due to higher fusion rates or some other perturbation, so will the pressure, and the star will expand. The temperature and density will soon decrease, lowering fusion rates. In turn, the pressure decreases, putting the star into a new stable (equilibrium) state. The same happens if the temperature drops.

Degenerate matter does not obey the ideal gas law. In objects mainly supported by electron degeneracy pressure, there is a different equation of state. The non-relativistic form is: $$P\propto\rho^{5/3}$$ while the relativistic form is $$P\propto\rho^{4/3}$$ Note the total lack of temperature-dependence. This means that when the temperature rises in a body supported by this pressure, there is nothing akin to the thermostat mechanism to lower the temperature. The results can be catastrophic. Type Ia supernovae, for instance, happen when a white dwarf accretes matter and begins fusing it on its surface. If the fusion rate is too high, the temperature rises, and runaway fusion may destroy the star.1

A sort of dance between the two relations is the mechanism behind a helium flash. In a star slightly more massive than the Sun ($\sim2$ solar masses), when hydrogen fusion ends in the core, the star shrinks a little, as fusion continues in the outer layers. Eventually, degenerate matter is formed in the core, and the inner part of the star is now supported by degeneracy pressure, rather than thermal pressure. Eventually, the core shrinks enough and gets so hot that helium quickly fuses into carbon, creating a seemingly runaway reaction. However, as the star expands again, thermal pressure in the outer layers takes over, calming the star. A large amount of energy is released (hence the name "helium flash"), but the star survives.

Your scenario is interesting because if the other side of the wormhole opens into any high-temperature, high-pressure environment, there will be absolutely disastrous consequences. The only hope for there not being an explosion would be if, for some reason, the wormholes are extremely short-lived most of the time, and so only a very small amount of material can travel through.

The semi-handwavy explanation you might be after is that the wormholes form because of some temperature differential. The Bose-Einstein condensate is cold, and the star-like object is hot. We could pretend that only high differences in temperature allow the wormholes to exist at all.

Also, assume when the wormholes are opened into whatever target object you're after, cause increases in temperature. At first, this isn't a problem, because the amount of matter the scientists are using isn't that large. After enough time, however, this effect builds up if it is not mitigated by the target body through cooling (which you could claim then closes the wormholes, allowing only a small amount of matter to go through), and an explosion occurs.

If you want to do something like this, then I would recommend using a white dwarf in a binary system that is already accreting matter. The wormholes simply provide the catalyst for fusion to begin, and there is no way to stop it. Higher temperatures mean a higher temperature difference. Thus, the wormholes stay open longer, and all of a sudden a significant amount of matter travels through - as the white dwarf undergoes a Type Ia supernova.

Side note: Neutron stars

One thing I realized I totally ignored was the equation of state for neutron stars. There are two reasons for this:

  1. It must be relativistic, meaning it's quite complicated.
  2. It's not actually well-known, and is an active area of study.

Neutron stars are massive and compact, and therefore an accurate treatment of them needs to use general relativity2 (although non-relativistic approximations could work to some degree; if you're really curious, you could look at polytropic models with indices of $0\leq n\leq1.5$). The other issue is that we don't yet know what equation of state best describes neutron stars. More observations (especially of binary systems and pulsars) are needed to place appropriate constraints, although several have already been proposed. Lattimer (2013) is a good review of various neutron star equations of state.

For this reason, I've decided not to discuss this option; we simply don't know enough to determine exactly what the effects would be in your scenario. I can tell you, though, that they would not be good for the physicists.

Other degenerate objects, like quark stars, also have complicated and not-yet-well-constrained equations of state. I talk about these in some more detail in another answer of mine.


1 This isn't guaranteed to happen; the shockwave may instead fizzle out, in what is called a deflagration wave (as opposed to a successful detonation wave). Modeling how these types of waves arise has been the target of simulations for quite some time.

2 Relativistic models are also quite nice for white dwarfs; see, for instance, this project.

$\endgroup$
  • 3
    $\begingroup$ If the very careful, cold, conditions caused the wormhole to form, then letting hot gas in will disrupt those conditions. The valve is only open for a small fraction of a second, but lets enough material through to cause an explosion in a vessel that’s designed to keep pressure on the other side of the wall. So it might not have much strength to keep pressure in. $\endgroup$ – JDługosz Jun 5 '17 at 14:05
  • $\begingroup$ @JDługosz That's a good point. I've come up with an alternate explanation for why the combination of hot and cold leads to stability (and eventually instability). $\endgroup$ – HDE 226868 Jun 5 '17 at 14:15
  • $\begingroup$ Additionally, the wormhole doesn't need to open in the star's interior. It could open farther out and just get a blast of corona or stellar wind. It would still be hot and fast moving and more dense than the vacuum chamber, but it wouldn't be overly excessive. $\endgroup$ – Draco18s Jun 5 '17 at 14:33
  • $\begingroup$ @Draco18s Maybe that would be the case for a normal star, but for a white dwarf, I doubt that would be the case. $\endgroup$ – HDE 226868 Jun 5 '17 at 16:02
  • $\begingroup$ The density of degenerate matter is controlled by how much pressure is being put on it. In white dwarves, electrons are no longer bound to atoms. In neutron stars, they have been literally smashed into protons, and only the Fermi exclusion principle keeps the nucleons from being crushed into each other. Releasing that sort of pressure, even for tiny amounts of matter would likely result in a devastating explosion as the nucleons raced apart at close to c. Investigators will be looking through the fine dust where the space station was to find out what happened. $\endgroup$ – Thucydides Jun 6 '17 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.