The cosmology of DIP

My first published short story, "Dip," has now been online at Aphelion for a few days (read it!). While the piece is written mostly as a character study--Nehal Gupta is someone ground to the point of breaking between the sublime and the mundane--the SF element of the story, the pocket universe, is an interesting toy to play with in itself so I thought I'd take a minute to describe what's going on under the surface.

Our universe is expanding in line with Einstein's theory of general relativity, if we take on board the idea that the universe is dominated by a mysterious "dark energy" that drives its expansion far more than the matter and energy we can see directly does. We don't know what the dark energy is, but relativity as a theory has held up to every test we've put it through.

The so-called Friedmann equations which describe how a universe expands can be written down like this:
(1) $\frac{\dot{a}^2 + k}{a^2} = \frac{\kappa \rho + \Lambda}{3}$
(2) $\frac{\ddot{a}}{a} = -\frac{1}{6}\left(1+3w\right) \kappa \rho + \frac{\Lambda}{3}$
(in this statement of the equation I've used a common physics convention of representing $8\pi$ times the gravitational constant with the letter $\kappa$ and using a set of units where the speed of light $c = 1$). What are the other letters doing?
  • a is the "scale factor," which tells us how much bigger (or smaller) the universe we're looking at is compared to some reference value. In the Friedmann-LemaĆ®tre-Robertson-Walker model of cosmology, the scale factor is a function only of time--the universe evolves the same all over.
  • The letter k describes the geometry of the universe: if k is 0 the universe is "flat" (like ours seems to be on large scales), if k is 1 the universe is closed and if k is -1 the universe is open.
  • $\rho$ is how much matter there is in a given volume of the universe. This is the easiest parameter for us to measure: it's simply how much stuff we see in the observable universe, divided by the volume of space we can observe.
  • $\Lambda$ is the so-called "cosmological constant." This is an energy field which is supposed to fill the universe and is built into space, regardless of the matter content. In "Dip," Fred Tarkies's battery-powered circuit uses electrical energy from our universe to create a cosmological constant in the pocket universe, via handwavy handwavy I don't know artistic license.
  • The little dots over the a mean rates of change. One dot over the a means, "how fast is the scale factor changing with time?" Two dots means, "how fast is the change in scale factor with time, itself, changing?"
  • Lastly we have w. w is what we call an "equation of state" and this basically talks about how the matter in the universe responds to being squeezed. Some illustrations: when we have $w=0$ this is a good model for the universe as we see it in real life: it is mostly empty space so we can squeeze pieces of it down at will. If $w=1$ meanwhile this is something that is so stiff it is incompressible, like a liquid or a rigid solid.
All this looks like a mess but we can break it down pretty easily if we just keep track of what each symbol is and what it's talking about, and also remember that the equations are on some level linear, that is, if we can solve them for a set of conditions that are almost right, we'll get an answer that's good enough to talk about what's going on qualitatively.

Case 1: a static pocket universe filled with guacamole

In this situation, we have a pocket universe that is held in an unchanging condition, i.e. it is static; and it is filled with a basically incompressible fluid (w is about 1), i.e. guacamole. This situation is easy to analyze because remember, the dotted terms represent rates of change and if something isn't changing, then the rate of change is zero so $\dot{a}=\ddot{a}=0$. So equation (1) above becomes
(3) $\frac{k}{a^2} = \frac{\kappa \rho + \Lambda}{3}$
while equation (2) becomes even simpler:
(4) $2\kappa \rho =  \Lambda.$
There are two key things to take away from this: first, our guacamole-filled universe needs to have k=1, that is, it's closed--otherwise the equations break down and we have no solution; and second, filling the universe with guacamole means the cosmological constant has to have a positive sign to balance it out and keep it static.

Fred's circuit makes sure the cosmological constant supplied to the pocket universe balances out the content of that universe by using electrical energy to create a positive $\Lambda$. (If this were to ever work in real life, the pocket universe would need to have a different "coupling constant" relating gravity to electrical charge, or else the energy requirements would be about $10^{17}$ joules for every kilogram of matter we put in the pocket universe--not even vaguely economical!)

Case 2: A runaway expanding pocket universe with very little guacamole

So what happens to Fred's beautiful circuit? As Nehal predicted, an oscillator--a circuit using negative feedback to regulate itself--always wants to become an amplifier, and in Nehal and Fred's case it has: the audience has eaten the guacamole faster than the circuit can handle so negative feedback turns into positive feedback, and Fred's circuit is now putting as much energy as it can into the cosmological constant. Let's model it this way: now a is changing with time and we have to solve for it, the universe is basically empty so $\rho = 0$ (in this case the equation of state stops mattering), and the cosmological constant is some large positive value. In this case equation (1) becomes
(5) $\frac{\dot{a}^2 + 1}{a^2} = \frac{\Lambda}{3}$ 
and yes indeed, we do have to cope with properly solving the differential equation. Luckily with a bit of algebra and separation of variables, we pretty quickly get an answer: the universe is expanding like the hyperbolic cosine function. If this is too heavy for you, don't worry: a good approximation for this is good old exponential growth, which gets faster the bigger the cosmological constant is (the scale factor goes like e to the power of time times the square root of $\Lambda$ divided by three, to be precise). This universe, if nothing changes, will keep expanding forever. When the batteries of Fred's circuit die, the influence of the matter that's been sucked in from our universe will take over and we end up with a universe that grows more or less like the one we live in today.

Fortunately, Nehal sees a solution.

Case 3: Solving the problem

What happens if we have a universe that runs like this but we change $\Lambda$ from positive to negative? Remember that the square root of a negative number has to involve the imaginary unit i. You might be tempted, then, to think that all the answers to our equations will be imaginary, but no: according to Swiss mathematician Leonhard Euler's famous identity for i in trigonometry, we will in fact get a real answer, something we can measure directly. In fact, it's exactly what we want: the universe which was growing exponentially in case 2, will now contract to a size of zero and vanish--and the more negative cosmological constant we give it, the faster that will happen. All Nehal has to do is reverse the two wires, red with black, and she can save the world...

Anyway. Read the thing. I'm supposed to be on vacation.

Edmund Schluessel

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