Undivided Looking

A reader asks:

This seems to be as good a place as any to ask a question about closed universes.

See, in a lot of popular science books, they teach you that an "open" universe is one where space is infinite, saddle-shaped, and keeps expanding forever; a "flat" universe is infinite, plane-shaped, and the rate of expansion eventually peters out to zero; and a "closed" universe is finite, sphere-shaped, and eventually contracts in a big crunch. They then talk about the cosmological constant and "dark energy," which make our universe expand at an accelerating rate, something that doesn't fit the taxonomy of possibilities for the universe's topology, and which they do not relate back to that taxonomy in any way.

Can a universe with lots of dark energy be a closed universe? Will a closed universe with dark energy keep on expanding and accelerating, or will it eventually collapse in a big crunch like a "normal" closed universe? Is the three-type Taxonomy only relevant given certain energy conditions? (Strong/weak/null)

Oh, and I almost forgot:

are there any good reasons to think that the universe is closed in the first place, other than Kalam-esqe arguments against actual infinities?

David,
It sounds like these books were just adding the new material about the cosmological constant to the old discussions without doing the hard work of going back and revising it so that it makes sense.

The Bad Old Days

In the old days (pre circa 1998) people didn't know about the acceleration of the universe, and they thought that the universe just consisted of ordinary radiation and matter (where for these purposes, dark matter is a form of matter).  In the old days, the model of closed, flat, and open works exactly as you say: a closed universe (spherical geometry) will recollapse, and open one (hyperbolic geometry) will trend to a constant rate of expansion (in terms of distance / time) and a flat one is right on the edge and will expand forever at a slower and slower rate (but still getting arbitrarily large).

Given the rate of expansion, it takes a certain amount of energy density to get a flat universe.  Too much, and you get a sphere, too little and you get hyperbolic space.  (The expansion or contraction of the universe makes it hyperbolic in the absence of matter.)  These are the 3 kinds of geometries which are homogeneous (the same everywhere) and isotropic (the same in every direction).  On average, the observable universe seems to be homogenous and isotropic, so it's got to be one of these three (a.k.a. an "FRW cosmology").

However, this was confusing for several reasons.  One is that the cosmological data kept suggesting that there wasn't enough energy in matter to get anywhere close to a flat universe, yet other data seemed more consistent with a flat universe.  A flat universe is also a natural consequence of inflation since it stretches out the pre-existing geometry to exponentially large distance scales.  Also, the universe seemed like it wasn't quite old enough to explain all the structures in it.

Concordance Cosmology

Now we know that there is an additional form of energy which is confusingly called "dark energy" (but I dislike this name, because it makes people think it has something to do with "dark matter".)  Most likely it is just a cosmological constant, a constant energy density in all of space.

Now it turns out that for purposes of determining the spatial geometry, a positive cosmological constant counts positively (so it helps to close the universe).  But when you calculate its effect on the expansion of the universe, it counts negatively, as repulsive gravity.

This may seem like odd behavior because energy and mass are equivalent, and we all know that mass causes gravitational attraction, not repulsion.  But in turns out that in General Relativity, both energy density (associated with time) and pressure (associated with space) lead to attractive gravity.  Negative pressure is called tension, and tension therefore causes antigravity.

In ordinary matter travelling at low speeds, the amount of pressure/tension is typically very small compared to the energy density.  Radiation which travels near the speed of light has a lot of pressure, but that only makes gravity stronger.

On the other hand, a positive cosmological constant has tension equal to its energy density.  Something has tension if, when you stretch it out, it's energy increases.  But the energy of the cosmological constant is proportional to the volume, so when the volume increases the energy increases proportionally.  Hence the tension in each spatial direction is equal to the energy.  Since there are 3 dimensions of space and only 1 of time, the antigravity due to the tension is 3 times larger than the gravity due to the energy density.  Hence the antigravity wins!  So paradoxically, the gravitational effects of this tension just make the universe want to grow faster!  Unlike the usual effects of tension, which cause things to shrink in on themselves.

On the other hand, if the cosmological constant were negative (it isn't, but suppose) its effects would be reversed: it would make the spatial geometry more hyperbolic, but would decelerate the expansion.

So, once you include a cosmological constant, the rules change (as you guessed).  You can still have the same 3 types of spatial geometry (the words "open", "flat", and "closed" describe the spatial geometry, not the dynamics).  But with a positive cosmological constant, even a universe with closed topology can sometimes expand forever, if it gets big enough for the cosmological constant to take over.  (Matter thins out, while the CC doesn't, so when the universe is small the matter is more important, and when it gets larger the CC is more important.)  On the other hand, with a negative cosmological constant, even an open cosmology will always eventually recollapse when it gets big enough.

(The various energy conditions you mention place limits on the allowed energy density and/or tension/pressure, so not surprisingly these have certain implications for what a cosmology can do.  Note that a positive CC violates the strong energy condition—which allows for a bounce, at least in the case of a closed universe.  While a negative CC violates the weak energy condition, which requires that any FRW cosmology which is neither expanding nor contracting at some time, must be closed.  (OK, technically it also allows space to be flat, but only if the matter energy is exactly 0, which is unrealistic.))

Our universe seems to have a positive cosmological constant, which fixes all of the problems I mentioned above.  The cosmological constant seems to give us exactly the extra energy density we need to get a flat universe.  Yet it also causes the universe to be currently accelerating in its expansion (lengthening the projected time back to the Big Bang); this acceleration has been confirmed by surveys of supernovae in the past.  So everything seems to hang together consistently.

As far as we can tell from current observation, the universe is exactly flat (with experimental error of about 1-2% over scales comparable to the observable universe)  However, a flat geometry is right on the knife's edge between the spherical and hyperbolic cases, so actually this is perfectly compatible with the universe having a tiny positive or negative curvature, as long as the radius of curvature is big enough.

So really it could still be any of the three cases, or else something more irregular.  As I said, inflation blows up the size of the universe, so regardless of the initial geometry, the observable universe will look flat after enough inflation.  Outside the observable universe, for all we know, it could be some other shape, perhaps it isn't even symmetrical.

There is really no particularly good physics reason, apart from aesthetics and philosophical bias to think that the universe should be closed or open.  I personally don't think much of the "Kalam" argument that actual infinities are impossible, but I do find it distasteful that in an infinite homogeneous universe everything (including all possible histories of the Earth) would happen infinitely many times in different places.

Also on the speculative hypothesis that the universe originated from some kind of quantum fluctuation, or no-boundary condition, I think one expects it to be closed.  But this kind of thing is extremely speculative.

If I had to place a bet with a metaphysical bookie, my money would be on closed (but enormously large so that we could never tell).  But this is my own personal guess, not a conclusion of Science!

(Incidentally, even if the topology of space is flat or hyperbolic, it would still be possible for the universe to be finite in size and therefore closed, so long as it has nontrivial topology.  For example, space could be a really big "torus" where if you go far enough in one direction, you come back around on the other side, like in some video games.  Locally, such a universe couldn't be distinguished from the infinite case, but globally it would be different.  Astronomers have done measurements looking for nontrivial topology in the sky.  They haven't seen anything, but of course they wouldn't if it happened on a scale much bigger than the observable universe!)

On the other hand, if the universe really does have a positive cosmological constant than (regardless of its spatial geometry) the final outcome seems secure.  If we extrapolate the current laws of physics to the far future (assuming no changes or interventions), we get an exponentially growing universe.  The matter thins out and becomes unimportant, and you end up with a very tiny final temperature (corresponding to the analogue of Hawking temperature but for cosmological horizons instead of black hole event horizons).

[Fixed typo in Schrodinger's equation below—AW]

Previously I talked about interference, the chief weird thing about QM that makes it different from Classical Mechanics.  You have to think about complex-valued "amplitudes", from which you derive (real-valued) probabilities.  From this you can also derive the notion of a Hilbert Space of states.  We discussed the space of states for the polarization of a photon (a 2 state system), and how there are many different choices of "basis", representing different ways of identifying a mutually exclusive set of two possibilities.

Now let's consider a more complicated system: a single particle moving around in empty space.  There are infinitely many states, because space is a continuum.  Hence we need to use an infinite dimensional Hilbert space.  This is harder to visualize than a two-dimensional one, but it will still be true that, in any given basis, each state can be regarded as a quantum superposition of a bunch of possibilities.

There are many possible choices of basis, but two of them are particularly nice.  You can choose to either express the system as superposition of position states, or as a superposition of momentum states, but you can't specify both at the same time, because they are two different bases of the Hilbert Space!  This is the origin of the Heisenberg Uncertainty Principle.

Of course, since the position and momentum are continuous variables, the probability of having any particular exact value of position or momentum is always 0.  So we have to generalize the framework slightly and talk about amplitude densities.  (However, there are other choices of basis where you don't have to do that).

An amplitude density is an amplitude per unit square-root-of-volume.  I know these units sound a bit strange, but that way when you square it, you get a probability per unit volume, which is as things should be for purposes of doing measurements.  The amplitude density is more commonly called the wavefunction of the particle.  So the wavefunction can be written as a function of position: $\Psi(x,\,y,\,z)$, or as a function of momentum: $\Psi(p_x,\,p_y,\,p_z)$, but not both at the same time.  However, if you know one of them, you can calculate the other one by a Fourier transform (should you be lucky enough to know what that is).

If you have multiple particles, you shouldn't think that each particle has a separate wavefunction.  Instead, you use a single wavefunction which depends on the positions (or momenta) of all of the particles.  For example, if there are two different particles which we'll call #1 and #2, then you'd write:

$$\Psi(\vec{r}_1, \vec{r}_2),$$

where I'm now using vector notation as a shorthand; but each vector still has x, y, and z components.  Hence the wavefunction for 2 particles actually is a function living in a 6 dimensional space!  (More generally, the wavefunction of $N$ particles will live in $3N$-dimensions, assuming that space is 3 dimensional.)

(Given these two particles, it might be that the wavefunction factorizes, so that

$$\Psi(\vec{r}_1, \vec{r}_2) = \Psi_1(\vec{r}_1)\Psi_2(\vec{r}_2).$$

That's what would happen if you independently prepare each particle in a state, and don't let them interact with each other.  But in general, there's lots of wavefunctions you could write down which do not factorize in this way.  This allows the particles to be correlated in strange ways not allowed by classical physics.  We call this phenomenon entanglement.)

Now the two particles might be either different type, or the same type.  One of the principles of particle physics is that apart from a limited number of attributes such as position/momentum, "spin", and a few other things, all particles of a given type are identical.  (E.g. all electrons have identical properties, and all photons also have identical properties.)

If the two particles are identical, then it shouldn't make any difference which particle we choose to label as "1" and which we choose to label as "2".  So there should be a symmetry of the wavefunction if we switch the two particles.  (Remember, in QM we have interference whenever two histories end up in the same place, so to get things right we have to obsess about exactly when two situations count as exactly the same, and when they don't.)  There are two different ways to implement this symmetry.  The obvious thing to do is to say that:

$$\Psi(\vec{r}_1, \vec{r}_2) = \Psi(\vec{r}_2, \vec{r}_1),$$

so that the amplitude is the same in both cases.  This sensible approach is taken by identical bosons, which includes particles such as photons, gluons, gravitons, mesons, He-4 nuclei, and so on.

Another, more perverse way to implement the symmetry is to insert a minus sign:

$$\Psi(\vec{r}_1, \vec{r}_2) = -\Psi(\vec{r}_2, \vec{r}_1).$$

This bizarre form of identicalness is used by fermions such as electrons, quarks, neutrinos, protons, neutrons, and He-3 nuclei.  (In general, something made out of an odd number of fermions is also a fermion, since if you switch two copies, you'll get an odd number of minus signs.)

So photons are strictly identical, while electrons are almost identical, but you get a minus sign if you switch them.  But remember, the overall phase of a QM system doesn't matter.  So you won't actually notice anything weird if you definitely switch two fermions.  The minus sign only matters in situations where they might-or-might-not have gotten switched, because then the interference between the two histories will be different.

A somewhat more straightforward implication is that no two identical fermions are ever in exactly the same position, because then the weird antisymmetry tells us that $\Psi(x_1, x_1) = - \Psi(x_1, x_1) = 0$.  This is a special case of the Pauli Exclusion Principle, which is the reason for the Periodic Table.  Since electrons can be either "spin up" or "spin down", you can only put 2 distinct electrons in each energy level of an atom.  Then the energy levels get full, and you have to put the electrons into higher energy shells.

Bosons, on the other hand, are gregarious and love to be in the same place.  Or rather, to speak less anthropomorphically, their probability to be in the same place is greater than you would expect from classical probability theory.  This is what makes lasers (a bunch of photons all in the same state) practically possible.

I've mentioned "spin" several times, but I haven't actually said what it is.  In QM, some particles also have an intrinsic angular momentum or polarization, which gives them a certain sort of directionality in space (even though they are point particles).  Unlike a lot of the cute terms used in particle physics such as "color" or "charm", the term "spin" really does refer to actual literal angular momentum.  But it works in a weird way.  The angular momentum along any axis is quantized, meaning it has to be either an integer or an integer + 1/2 (times the Planck constant $\hbar$).  The maximum possible angular momentum along any axis is called the "spin" of the particle.

In Nature, there is a rule called spin-statistics which says that particles with integer spin are always bosons, and particles with half-integer spin are always fermions.  (You can prove this rule mathematically in QFT, but it requires Special Relativity and some additional physical assumptions.)

Every known fundamental fermion is spin 1/2, which means that along any given axis it is either spinning clockwise (a.k.a. "up" or +1/2) or counterclockwise (a.k.a. "down" or -1/2).  You only get to specify the spin along one axis, say the vertical one.  This is not to say that an electron can't spin "left", "right", "in", or "out", but these states are quantum superpositions of the "up" and "down" states.  By rotational symmetry, we could pick a different basis (e.g. right / left) and instead think of up and down as superpositions of right and left.  The (2 complex dimensions = 4 real dimensions) space of possible electron spins is called a spinor.

A spinor needs to be rotated by 720º (2 full circles) to get back to its original state.  Yes, you read that right.  If you only rotate it by 360º (1 full circle) then it comes back to itself with an extra minus sign in the amplitude.  Just like when you switch two electrons.  They're just perverse that way.

Most of the fundamental bosons are spin 1, so their polarization is given by a vector, as in the case of the photon which we discussed last time.  Vectors get a minus sign when you rotate them by 180º, and return back to the way they were after 360º, just like you were taught in school.

However, the Higgs boson (which gives mass to most of the other fundamental particles) is a spin-0 or scalar field.  That means it doesn't change at all when you rotate it. On the other hand, the graviton is a spin-2 particle, which means it gets a minus sign when you rotate it 90º, and goes to itself under 180º.  Its polarization is described by a matrix, but let's not get into that here.

The bottom line is that for anything more complicated than a scalar field, in addition to the position or momentum variables you also need to include the spin degrees of freedom.  So if we have one electron and one photon, the wavefunction will look like e.g.

$$\Psi(\vec{p}_e, s_e,\vec{p}_\gamma, s_\gamma),$$

where $s$ represents the spin of the electron or photon along e.g. the z-axis.  This is still a 6 dimensional space since $s_e$ can only take the values $(+1/2, -1/2)$, and $s_\gamma$ can only take the values $(+1,0,-1)$.  (Incidentally, since the photon is massless, its spin is always required to be perpendicular to its momentum, so there are really only 2 polarization states, not 3.  But the explanation of this involves relativity and gauge symmetry and a bunch of other things from QFT...)

There is also a third kind of basis, distinct in general from both the position and the momentum basis, in which time evolution is particularly simple.  This is the basis where the energy of the system takes on a definite value.  In this basis, the only thing that changes is the phase of each energy state.  The phase changes with time at a speed proportional to the energy.

So one way to specify the dynamics of a QM system is simply to say what the formula for the energy $H$ is, as a function of all the positions and momenta of all the particles in the problem.  (You think of this an operator, a gadget which acts on the wavefunction to get another wavefunction.  So if you are in the position basis, the "momentum operator" is given by $\vec{p} \Psi = (i / \hbar) \vec{\nabla} \Psi$, which is equivalent to switching from the position to the momentum basis, multiplying by p, and then switching back.)  Then you can figure out how the wavefunction changes with time by using the Schrodinger equation:

$$\frac{d\Psi}{dt} = -\frac{i}{\hbar} H \Psi.$$

Thus, if you know what the formula for the energy is, you can predict the dynamics of the wavefunction as time passes.  This is related to the Hamiltonian approach in Classical Mechanics.  As you take the $\hbar \to 0$ limit, you recover classical mechanics.

There is also a "path integral" picture due to Feynman, related to the Lagrangian or "Least Action" approach to physics mentioned at the same link, where you assign to each history an amplitude proportional to $e^{iS/\hbar}$, where $S$ is the action.  This approach is actually more closely related to the picture I started with in part I!

In this sense, Quantum Mechanics is a fulfillment of Classical Mechanics, just as (in Christian doctrine) the New Covenant fulfills the Old Covenant.  That is, the new model justifies the quirky, previously-inexplicable features of the old model, in terms of a more basic (yet also more mysterious) set of ideas.  Concepts such as action, energy, momentum, the associated conservation laws, and so on, all follow naturally from the interference of wavefunctions over space and time.

Incidentally, there's a very important flaw in what I've told you so far.  Generally speaking, in modern physics it's better to think of the universe as being made of fields, not particles!  This is the subject of Quantum Field Theory.  The idea is that we should really think of the universe as being made of some finite number of types of fields (e.g. the electron field, the photon/EM field, the quark field, etc.).  Consider a scalar field $\Phi(t,x,y,z)$, which is basically a function of the spacetime points.  If we want to keep track of the amplitude for any possible configuration of the field, then we really need our wavefunction to be a function of all possible configurations of $\Phi$ at one moment of time.  Morally speaking (i.e. I am about to make certain dreadful oversimplifications) this means that the state of the universe at one time is something more like:

$$\Psi(\Phi(\vec{r})),$$

in other words the wavefunction is a function of functions!  The "particles" are then quantized excitations associated with different modes of this field.  (The relationship between QFT and the QM of multiple particles should not be obvious from what I've said so far...)

QFT gets kind of complicated, but the advantages are that 1) it is easier to make it compatible with Special Relativity, and 2) it allows one to consider situations where particles are created and destroyed, e.g. an electron can emit or absorb a photon.  Since this happens all the time in the real world, that's kind of important!

But as long as you're dealing with situations where the particles are all going much slower than the speed of light, and none of them decay into other particles, you can use QM as described above.  (Perhaps I shouldn't have given the photon as an example, because it always travels at the speed of light and is never nonrelativistic.)