# February 20, 1993 {#week6}

1) Alexander Vilenkin, _Quantum cosmology_, talk given at Texas/Pascos
1992 at Berkeley, available as [`gr-qc/9302016`](https://arxiv.org/abs/gr-qc/9302016).

This is, as Vilenkin notes, an elementary review of quantum cosmology.
It won't be news to anyone who has kept up on that subject (except
perhaps for a few speculations at the end), but for those who haven't
been following this stuff, like myself, it might be a good way to get
started.

Let's get warmed up....

Quantizing gravity is mighty hard. For one thing, there's the "problem
of time" --- the lack of a distinguished time parameter in *classical*
general relativity means that the usual recipe for quantizing a
dynamical system --- "represent time evolution by the unitary operators
exp(-iHt) on the Hilbert space of states, where t is the time and H, the
Hamiltonian, is a self-adjoint operator" --- breaks down! As Wheeler so
picturesquely put it, in general relativity we have "many-fingered
time"; there are lots of ways of pushing a spacelike surface forwards
in time.

But if we simplify the heck out of the problem, we might make a little
progress. (This is a standard method in physics, and whether or not
it's really justified, it's often the only thing one can do!) For one
thing, note that in the big bang cosmology there is a distinguished
"rest frame" (or more precisely, field of timelike vectors) given by
the galaxies, if we discount their small random motions. In reality
these are maybe not so small, and maybe not so random --- such things as
the "Virgo flow" show this --- but we're talking strictly theory here,
okay? --- so don't bother us with facts! So, if we imagine that things go
the way the simplest big bang models predict, the galaxies just sit
there like dots on a balloon that is being inflated, defining a notion
of "rest" at each point in spacetime. This gives a corresponding
notion of time, since one can measure time using clocks that are at rest
relative to the galaxies. Then, since we are pretending the universe is
completely homogeneous and isotropic --- and let's say it's a closed
universe in the shape of a 3-sphere, to be specific --- the metric is
given by
$$dt^2-r(t)^2[(d\psi)^2 + (\sin \psi)^2{(d\theta)^2 + (\sin \theta)^2 (d\varphi)^2}]$$
What does all this mean? Here $r(t)$ is the radius of the universe as a
function of time, the following stuff is just the usual metric on the
unit 3-sphere with hyperspherical coordinates $\psi$, $\theta$, $\varphi$ generalizing the
standard coordinates on the 2-sphere we all learn in college:
$$(d\psi)^2 + (\sin \psi)^2{(d\theta)^2 + (\sin \theta)^2 (d\varphi)^2}$$
and the fact that the metric on spacetime is $dt^2$ minus a bunch of
stuff reflects the fact that spacetime geometry is "Lorentzian," just
as in flat Minkowski space the metric is
$$dt^2-dx^2-dy^2-dz^2.$$
The name of the game in this simple sort of big bang cosmology is thus
finding the function $r(t)$! To do this, of course, we need to see what
Einstein's equations reduce to in this special case, and since
Einstein's equations tell us how spacetime curves in response to the
stress-energy tensor, this will depend on what sort of matter we have
around. We are assuming that it's homogeneous and isotropic, whatever
it is, so it turns out that all we need to know is its density $\rho$ and
pressure $P$ (which are functions of time). We get the equations
$$\begin{aligned}r''/r &= -\frac{4\pi}{3}(\rho+3P) \\ (r')^2 &= \frac{8\pi}{3}\rho r^2-1\end{aligned}$$
Here primes denote differentiation with respect to $t$, and I'm using
units in which the gravitational constant and speed of light are equal
to 1.

Let's simplify this even more. Let's assume our matter is "dust,"
which is the technical term for zero pressure. We get two equations:
$$\begin{aligned}r''/r &= -\frac{4\pi}{3}\rho \\ (r')^2 &= \frac{8\pi}{3}\rho r^2-1\end{aligned}\tag{1}$$
Now let's take the second one, differentiate with respect to $t$,
$$2r''r' = \frac{8\pi}{3}(\rho'r^2 + 2 \rho r r')$$
plug in what the first equation said about r'',
$$-\frac{8\pi}{3} \rho r r' = \frac{8\pi}{3}(\rho' r^2 + 2 \rho r r')$$
clear out the crud, and lo:
$$3 \rho r' = -\rho' r$$
or, more enlighteningly,
$$\frac{d(\rho r^3)}{dt} = 0.$$
This is just "conservation of dust" --- the dust density times the
volume of the universe is staying constant. This, by the way, is a
special case of the fact that Einstein's equations *automatically
imply* local conservation of energy (i.e., that the stress-energy tensor
is divergence-free).

Okay, so let's say $\rho r^3 = D$, with $D$ being the total amount of dust.
Then we can eliminate $\rho$ from equations (1) and get:
$$\begin{aligned}r'' &= -\frac{4\pi D}{3r^2} \\ (r')^2-\frac{8\pi}{3}\frac{D}{r} &= -1 \end{aligned}\tag{2}$$
What does this mean? Well, the first one looks like it's saying
there's a force trying to make the universe collapse, and that the
strength of this force is proportional to $1/r^2$. Sound vaguely
familiar? It's actually misleadingly simple --- if we had put in
something besides dust it wouldn't work quite this way --- but as long as
we don't take it too seriously, we can just think of this as gravity
trying to get the universe to collapse. And the second one looks like
it's saying that the "kinetic" energy proportional to $(r')^2$, plus
the "potential" energy proportional to $-1/r$, is constant! In other
words, we have a nice analogy between the big bang cosmology and a very
old-fashioned system, a classical particle in one dimension attracted to
the origin by a $1/r^2$ force!

It's easy enough to solve this equation, and easier still to figure it
out qualitatively. The key thing is that since the total "energy" in
the second equation of (2) is negative, there won't be enough
"energy" for $r$ to go to infinity, that is, there'll be a big bang and
then a big crunch. Here's $r$ as a function of $t$, roughly:
$$
  \begin{tikzpicture}
    \draw[->] (0,0) to (5,0) node[label=below left:{$t$}]{};
    \draw[->] (0,0) to (0,2.5) node[label=below left:{$r$}]{};
    \draw[thick] plot [smooth,tension=1] coordinates {(0,0) (2,1.5) (4,0)};
  \end{tikzpicture}
  \text{Figure 1}
$$
What goes up, must come down! This curve, which I haven't drawn too
well, is just a cycloid, which is the curve traced out by a point on the
rim of rolling wheel. So, succumbing to romanticism momentarily we could
call this picture ONE TURN OF THE GREAT WHEEL OF TIME.... But there is
*no* reason to expect further turns, because the differential equation
simply becomes singular when $r = 0$. We may either say it doesn't make
sense to speak of "before the big bang" or "after the big crunch" ---
or we can look for improved laws that avoid these singularities. (I
should repeat that we are dealing with unrealistic models here, since
for example there is no evidence that there is enough matter around to
"close the universe" and make this solution qualitatively valid --- it
may well be that there's a big bang but no big crunch. In this case,
there's only one singularity to worry about, not two.)

People have certainly not been too ashamed to study the *quantum* theory
of this system (and souped-up variants) in an effort to get a little
insight into quantum gravity. We would expect that quantum effects
wouldn't matter much until the radius of the universe is very small,
but when it *is* very small they would matter a lot, and maybe --- one
might hope --- they would save the day, preventing the nasty
singularities. I'm not saying they DO --- this is hotly debated --- but
certainly some people hope they do. Of course, serious quantum gravity
should take into account the fact that geometry of spacetime has all
sorts of wiggles in it -it isn't just a symmetrical sphere. This may
make a vast difference in how things work out. (For example, the big
crunch would be a lot more exciting if there were lots of black holes
around by then.) The technical term for the space of all metrics on
space is "superspace" (sigh), and the toy models one gets by ignoring
all but finitely many degrees of freedom are called "minisuperspace"
models.

Let's look at a simple minisuperspace model. The simplest thing to try
is to take the classical equations of motion (2) and try to quantize
them just like one would a particle in a potential. This is a delicate
business, by the way, because one can't just take some classical
equations of motion and quantize them in any routine way. There are lots
of methods of quantization, but all of them require a certain amount of
case-by-case finesse.

The idea of "canonical quantization" of a classical system with one
degree of freedom --- like our big bang model above, where the one degree
of freedom is $r$ --- is to turn the "position" (that's $r$) into a
multiplication operator and the "momentum" (often that's something
like $r'$, but watch out!) into a differentation operator, say $-i \hbar \frac{d}{dr}$,
so that we get the "canonical commutation relations"
$$[-i \hbar \frac{d}{dr}, r] = -i \hbar.$$
We then take the formula for the energy, or Hamiltonian, in terms of
position and momentum, and plug in these operators, so that the
Hamiltonian becomes an operator. (Here various "operator-ordering"
problems can arise, because the position and momentum commuted in the
original classical system but not anymore!) To explain what I mean, why
don't I just do it!

So: I said that the formula
$$(r')^2-\frac{8\pi}{3} \frac{D}{r} = -1 \tag{3}$$
looks a lot like a formula of the form "kinetic energy plus potential
energy is constant". Of course, we could multiply the whole equation by
anything and get a valid equation, so it's not obvious that the
"right" Hamiltonian is
$$(r')^2-\frac{8\pi}{3} \frac{D}{r}$$
or (adding 1 doesn't hurt)
$$(r')^2-\frac{8\pi}{3} \frac{D}{r} + 1$$
In fact, note that multiplying the Hamiltonian by some function of $r$
just amounts to reparametrizing time, which is perfectly fine in general
relativity. In fact, Vilenkin and other before him have decided it's
better to multiply the Hamiltonian above by $r^2$. Why? Well, it has to
do with figuring out what the right notion of "momentum" is
corresponding to the "position" $r$. Let's do that. We use the old
formula
$$p = \frac{dL}{dq'}$$
relating momentum to the Lagrangian, where for us the position, usually
called $q$, is really $r$.

The Lagrangian of general relativity is the "Ricci scalar" $R$ --- a
measure of curvature of the metric --- and in the present problem it turns
out to be
$$R = 6 \left(\frac{r''}{r} + \frac{(r')^2}{r^2}\right)$$
But we are reducing the full field theory problem down to a problem with
one degree of freedom, so our Lagrangian should be the above integrated
over the 3-sphere, which has volume $16 \pi r^3/3$, giving us
$$32\pi (r''r^2 + (r')^2 r)$$
However, the $r''$ is a nuisance, and we only use the integral of the
Lagrangian with respect to time (that's the action, which classically
is extremized to get the equations of motion), so let's do an
integration by parts, or in other words add a total divergence, to get
the Lagrangian
$$L = -32\pi (r')^2 r.$$
Differentiating with respect to $r'$ we get the momentum "conjugate to
$r$",
$$p = -64\pi r'r.$$
Now I notice that Vilenkin uses as the momentum simply $-r'r$, somehow
sweeping the monstrous $64\pi$ under the rug. I have the feeling that this
amounts to pushing this factor into the definition of $\hbar$ in the canonical
commutation relations. Since I was going to set $\hbar$ to 1 in a minute
anyway, this is okay (honest). So let's keep life simple and use
$$p = -r'r.$$
Okay! Now here's the point, we want to exploit the analogy with good
old quantum mechanics, which typically has Hamiltonians containing
something like $p^2$. So let's take our preliminary Hamiltonian
$$(r')^2-\frac{8\pi}{3} \frac{D}{r} + 1$$
and multiply it by $r^2$, getting
$$H = p^2-\frac{8\pi D}{3}r + r^2.$$
Hey, what's this? A harmonic oscillator! (Slightly shifted by the term
proportional to $r$.) So the universe is just a harmonic oscillator... I
guess that's why they stressed that so much in all my classes!

Actually, despite the fact that we are working with a very simple model
of quantum cosmology, it's not quite *that* simple. First of all,
recall our original classical equation, (3). This constrained the energy
to have a certain value. I.e., we are dealing not with a Hamiltonian in
the ordinary sense, but a "Hamiltonian constraint" --- typical of
systems with time reparametrization invariance. So our quantized
equation says that the "wavefunction of the universe," $\psi(r)$, must
satisfy
$$H \psi = 0.$$
Also, unlike the ordinary harmonic oscillator we have the requirement
that $r\geqslant0$. In other word, we're working with a problem that's like a
harmonic oscillator and a "wall" that keeps $r\geqslant0$. Think of a
particle in a potential like this:
$$
  \begin{tikzpicture}
    \draw[->] (0,0) to (5,0) node [label=below left:{$r$}]{};
    \draw[->] (0,-0.7) to (0,2.5) node[label=below left:{$V(r)$}]{};
    \draw[thick] plot [smooth,tension=1] coordinates {(0,0) (1.8,-0.5) (4,2.5)};
  \end{tikzpicture}
  \text{Figure 2}
$$
Here $V(r) = -(8\pi D/3)r + r^2$. The minimum of $V$ is at $r = 4 \pi D/3$ and
the zeroes are at $r = 0$ and $8 \pi D/3$. Classically, a particle with zero
energy starting at $r = 0$ will roll to the right and make it out to $r = 4\pi D/3$ before rolling back to $r = 0$. This is basically the picture we had
in Figure 1, except that we've reparametrized time so we have simple
harmonic motion instead of cycloid.

Quantum mechanically, however one must pick boundary conditions at $r = 0$
to make the problem well-defined!

This is where the fur begins to fly!! Hawking and Vilenkin have very
different ideas about what the right boundary conditions are. And note
that this is not a mere technical issue, since they determine the
wavefunction of the universe in this approach! I will not discuss this
since Vilenkin does so quite clearly, and if you understand what I have
written above you'll be in a decent position to understand him. I will
just note that Vilenkin, rather than working with a universe full of
"dust," considers a universe in which the dominant contribution to the
stress-energy tensor is the cosmological constant, that is, the negative
energy density of a "false vacuum", which believers in inflation (such
as Vilenkin) think powered the exponential growth of the universe at an
early stage. So his equations are slightly different from those above
(and are only meant to apply to the early history of the universe).

[Let me just interject a question to the experts if I may --- since I've
written this long article primarily to educate myself. It would seem to
me that the equation $H \psi = 0$ above would only have a normalizable
solution if the boundary conditions were fine-tuned! I.e., maybe the
equation $H \psi = 0$ itself determines the boundary conditions! This would
be very nice; has anyone thought of this? It seems reasonable because,
with typical boundary conditions, the operator $H$ above will have pure
point spectrum (only eigenvalues) and it would be rather special for one
of them to be $0$, allowing a normalizable solution of $H \psi = 0$. Also,
corrections and education of any sort are welcomed. I would love to
discuss this with some experts.]

Anyway, suppose we find some boundary conditions and calculate $\psi$, the
"wavefunction of the universe." (I like repeating that phrase because
it sounds so momentous, despite the fact that we are working with a
laughably oversimplified toy model.) What then? What are the
implications for the man in the street?

Let me get quite vague at this point. Think of the radius of the
universe as analogous to a particle moving in the potential of Figure 2.
In the current state of affairs classical mechanics is an excellent
approximation, so it seems to trace out a classical trajectory. Of
course it is really obeying the laws of quantum mechanics, so the
trajectory is really a "wave packet" --- technically, we use the WKB
approximation to see how the wave packet can seem like a classical
trajectory. But near the big bang or big crunch, quantum mechanics
matters a lot: there the potential is rapidly varying (in our simple
model it just becomes a "wall") and the wave packet may smear out
noticeably. (Think of how when you shoot an electron at a nucleus it
bounces off in an unpredictable direction --- it's wavefunction just
tells you the *probability* that it'll go this way or that!) So some
quantum cosmologists have suggested that if there is a big crunch, the
universe will pop back out in a highly unpredictable, random kind of
way!

I should note that Vilenkin has a very different picture. Since this
stuff makes large numbers of assumptions with very little supporting
evidence, it is science that's just on the brink of being mythology.
Still, it's very interesting.

2) Lee Smolin, "Finite, diffeomorphism invariant observables in quantum
gravity", available as [`gr-qc/9302011`](https://arxiv.org/abs/gr-qc/9302011).

The big problem in canonical quantization of gravity, once one gets
beyond "mini-" and "midisuperspace" models, is to find enough
diffeomorphism-invariant observables. There is a certain amount of
argument about this stuff, and various approaches, but one common
viewpoint is that the "physical" observables, that is, the really
observable observables, in general relativity are those that are
invariant under all diffeomorphisms of spacetime. I.e., those that are
independent of any choice of coordinates. For example, saying "My
position is $(242,2361,12,-17)$" is not diffeomorphism-invariant, but
saying "I'm having the time of my life" is. It's hard to find lots
of (tractable) diffeomorphism invariant observables --- or even any! Try
figuring out how you would precisely describe the shape of a rock
without introducing any coordinates, and you'll begin to see the
problem. (The quantum mechanical aspects make it harder.)

Rovelli came up a while back with a very clever angle on this problem.
It's rather artificial but still a big start. Using a "field of
clocks" he was able to come up with interesting diffeomorphism
invariant observables. The idea is simply that if you had clocks all
around you could say "when the bells rang 2 a.m. I was having the time
of my life" --- and this would be a diffeomorphism-invariant statement,
since rather than referring to an abstract coordinate system it
expresses the coincidence of two physical occurences, just like "the
baseball broke through the window". Then he pushed this idea to define
"evolving constants of motion" --- a deliberate oxymoron --- to deal with
the famous "problem of time" in general relativity: how to treat time
evolution in a coordinate-free manner on a spacetime that's not flat
and, worse, whose geometry is "uncertain" a la Heisenberg? This is
treated, by the way, in

3) Carlo Rovelli, "Time in quantum gravity: an hypothesis", _Phys. Rev._ **D43** (1991), 442--456.

Also, an excellent and very thorough review of the problem of time and
various proposed solutions, including Rovelli's, is given in

4) Chris J. Isham, "Canonical quantum gravity and the problem of time", 125 pages, available as [`gr-qc/9210011`](https://arxiv.org/abs/gr-qc/9210011).

Anyway, in a paper I very briefly described in ["Week 1"](#week1):

5) Lee Smolin, "Time, measurement and information loss in quantum
cosmology", available as
[`gr-qc/9301016`](https://arxiv.org/abs/gr-qc/9301016).

Smolin showed, how, using a clever trick sketched in the present paper
to get "observables" invariant under spatial but not temporal
observables, together with Rovelli's idea, one could define lots of
REAL observables, invariant under spacetime diffeomorphisms that is,
thus making a serious bite into this problem.

I warn the reader that there is a fair amount that is not too realistic
about these methods. First there's the "clock field" --- this can
actually be taken as a free massless scalar field, but in so doing there
is the likelihood of serious technical problems. Some of these are
discussed in

6) P. Hajicek, 'Comment on "Time in quantum gravity --- an hypothesis"',
_Phys. Rev._ **D44** (1991), 1337--1338.

(But I haven't actually read this, just Isham's description.) Also,
the clever trick of the present paper is to couple gravity to an
antisymmetric tensor gauge field so that in addition to having loops as
part of ones "loop representation," one has surfaces --- a "surface
representation". But this antisymmetric tensor gauge field is not the
sort of thing that actually seems to arise in physics (unless I'm
missing something). Still, it's a start. I think I'll finish by
quoting Smolin's abstract:

> Two sets of spatially diffeomorphism invariant operators are
> constructed in the loop representation formulation of quantum gravity.
> This is done by coupling general relativity to an anti- symmetric
> tensor gauge field and using that field to pick out sets of surfaces,
> with boundaries, in the spatial three manifold. The two sets of
> observables then measure the areas of these surfaces and the Wilson
> loops for the self-dual connection around their boundaries. The
> operators that represent these observables are finite and background
> independent when constructed through a proper regularization
> procedure. Furthermore, the spectra of the area operators are discrete
> so that the possible values that one can obtain by a measurement of
> the area of a physical surface in quantum gravity are valued in a
> discrete set that includes integral multiples of half the Planck area.
> These results make possible the construction of a correspondence
> between any three geometry whose curvature is small in Planck units
> and a diffeomorphism invariant state of the gravitational and matter
> fields. This correspondence relies on the approximation of the
> classical geometry by a piecewise flat Regge manifold, which is then
> put in correspondence with a diffeomorphism invariant state of the
> gravity-matter system in which the matter fields specify the faces of
> the triangulation and the gravitational field is in an eigenstate of
> the operators that measure their areas.

------------------------------------------------------------------------

> *In the Space and Time marriage we have the greatest Boy meets Girl
story of the age. To our great-grandchildren this will be as poetical a
union as the ancient Greek marriage of Cupid and Psyche seems to us.*
> 
> --- Lawrence Durrell