Why is quantum gravity hard?

October 2, 2020. People often say that quantum gravity is hard because quantum mechanics and gravity are incompatible. I’ll give a brief, non-technical explanation of the real problem: powerful microscopes make black holes. In an appendix, we’ll also see why there is no problem combining gravity and quantum mechanics as long as we stick to sufficiently weak microscopes.


Gravity is a theory of heavy things; quantum mechanics is a theory of fuzzy things; quantum gravity applies to things which are both heavy and fuzzy. People often say that the two theories are “incompatible” because they use different mathematical frameworks. This makes it seem like a technical problem, which is misleading. The real problem is that the details of quantum gravity are hidden inside black holes! Let’s see why.

Microscopes make black holes

First, we’ll make a heuristic argument that sufficiently powerful microscopes create black holes. It is really just a syllogism: microscopes make energy (by virtue of Heisenberg’s uncertainty principle), energy makes black holes (using $E = mc^2$ and gravity), hence microscopes make black holes.

Microscopes make energy

Suppose we have a microscope which can resolve lengths $\Delta x$. Heisenberg’s uncertainty principle says that the smaller this resolution, the larger the uncertainty about the momentum of things we measure, with

\[\Delta p \gtrsim \frac{\hbar}{\Delta x}\]

for Planck’s constant $\hbar \approx 10^{-34} \text{ J/s}$. We can relate this to the energy of the particles using Einstein’s famous $E = mc^2$. In fact, we will write it in the much less well-known form

\[E^2 = m^2c^4 = p^2c^2 + m_0^2 c^4,\]

where $m_0$ is the mass of the particle at rest, $c = 3 \times 10^8 \text{ m/s}$ is the speed of light, and $m$ is the relativistic mass, which increases (without limit) as the particle speeds up. When the particle is moving very quickly, the momentum can be much larger than the rest mass energy, and $p \approx E/c$. If we measure very small distances, Heisenberg’s principle tells us we will be smacking around particles at very high momenta, so this is the form we should use. Thus, the uncertainty in the energy of particles our microscope is examining is

\[\Delta E \sim \frac{\hbar c}{\Delta x}\;.\]
Energy makes black holes

Let’s now recall Newton’s universal law of gravitation,

\[F = \frac{Gm_1 m_2}{r^2},\]

where $G = 6.7\times 10^{-11} \text{ N m}^2\text{ /kg}$ is Newton’s constant. We can use this to estimate the size of a black hole! A black hole is a region of space where gravity is so strong light is unable to escape. To see how light figures in Newton’s law, we need to give it a mass. In classical physics, light is massless, but but we know better: Einstein’s formula tells us that it has some relativistic mass related to its energy, $E = mc^2$. [If you’re curious, the mass of a particle of light, the photon, depends on its frequency $f$ via

\[m = \frac{E}{c^2} = \frac{2\pi\hbar f}{c^2},\]

using the formula for the energy of a photon, $E = 2\pi \hbar f$, also discovered by Einstein.] Let’s continue. For a black hole of mass $M$, and radius $r_s$, the force it exerts on a photon with mass-energy $m$ is

\[F \sim \frac{GMm}{r_s^2}.\]

Using the work formula, we can view force as energy divided by distance. Since the energy of the photon is $E = mc^2$, and the relevant distance is probably the black hole size $r_s$, we have

\[\frac{mc^2}{r_s} \sim \frac{GMm}{r_s^2} \quad \Longrightarrow \quad r_s \sim \frac{GM}{c^2}.\]

Although we’ve been rather sloppy, this guess is correct up to a factor of $2$! So, if we take a mass $M$ and squish into a ball of radius $\lesssim GM/c^2$, it will make a black hole.

Concluding the syllogism

Let’s return to our microscope. Zooming in makes energy fluctuations, and by $E = mc^2$ these fluctuations have mass. The associated black hole radius is

\[r_s \sim \frac{Gm}{c^2} = \frac{G\Delta E}{c^4} = \frac{G\hbar}{\Delta x \cdot c^3}.\]

It may seem sketchy to replace $E$ with $\Delta E$, but if the energy of particles has fluctuations of size $\Delta E$ around $E = 0$, some of them will have energy $E$. We can clean up this expression by packaging all these constants into a single object called the Planck length:

\[\ell_P := \sqrt{\frac{G\hbar}{c^3}}.\]

Then the associated black hole radius for our microscope is

\[r_s \sim \frac{\ell_P^2}{\Delta x}.\]

Now, what does this all mean? Very simply, if the resolution $\Delta x$ of our microscope is within the associated black hole radius, then our energy fluctuations will produce a tiny black hole! We won’t see anything at all. This happens when

\[\Delta x \lesssim r_s \sim \frac{\ell_P^2}{\Delta x} \quad \Longrightarrow \quad \Delta x \lesssim \ell_P.\]

So, prying below the Planck scale makes black holes. Clearly, microscopes this powerful should be kept away from theoretical physicists!

Local observables and the shape of space

The basic difficulty with quantum gravity is that zooming in makes black holes. This problem appears in many different guises. We will discuss two here: the notion that local observables make no sense in a theory of quantum gravity, and Einstein’s observation that physical statements should only depend on the shape of space, not on how we choose to label it.

No local observables

Suppose I want to make statements about what is happening at a point $X$ in spacetime, e.g. an electron has some probability of being observed there. So, I wait around with my electron detector to see if the electron appears at $P$, and in due course, it does. Or does it really? If my electron detector can tell what’s happening at the point $P$ itself, it’s acting as a microscope with infinite resolution! Clearly, this is not consistent with the argument above. At best, I can look for electrons in regions larger than the Planck length.

Instead of talking about black holes, we usually say that spacetime breaks down when we try to look too close. For this reason, it doesn’t make sense to talk about what is happening at a point in spacetime. The electron detector is an example of something called a local observable: it is defined at a point (local) and makes a quantum-mechanical measurement (observable). This conclusion is so important we put it in a quote box:

In quantum gravity, there are no local observables.

In fact, this conclusion is already suggested by Einstein’s classical theory of gravity, general relativity. Without going into details, we can briefly explain why, and some of the fancy words physicists use to dress up the idea.

The shape of things

In Newtonian gravity, spacetime is a sort of fixed arena where objects float around and exert forces on each other. In this picture, gravity is just another force. Einstein’s profound realization was that gravity isn’t a real force at all! Rather, gravity is the shape of space, and space is shaped by matter. In the beautiful maxim of John Wheeler,

Spacetime tells matter how to move; matter tells spacetime how to curve.

It doesn’t make much sense to talk about stuff happening at a point $P$ itself. Rather, physically meaningful statements will involve the shape of space and matter in the vicinity of $P$. The shape of space is captured by an object $\mathbf{G}$ called the Einstein tensor, while the “shape” of matter is captured by an energy-like quantity called the stress-energy tensor $\mathbf{T}$. The governing equation of general relativity is the Einstein field equation:

\[\mathbf{G} = \frac{8\pi G}{c^4}\mathbf{T}.\]

Spacetime curves on the left; matter moves on the right; the equality sign means they are giving each other instructions. This is not the only theory of gravity you can write down, and indeed, there are many ways to generalize Einstein’s equation. But they should obey Einstein’s insight that gravitational physics only depends on the shape of things.

Physicists have a nice hack for enforcing this shape-only dependence. Clearly, the label of the point $P$ will depend on my labelling system. But however I choose to label $P$ (and the surrounding points), the shape of space doesn’t change. A sphere is still a sphere, even if I label points using fruits or days of the week! Thus, physically meaningful statements do not depend on labelling. Physicists like fancy terminology, and “relabelling points” often goes by “diffeomorphism”, and “does not depend on” by “invariant”. So in fancy language, the constraint becomes another quote:

Physically meaningful statements are invariant under diffeomorphisms.

I can’t really ask about whether an electron appeared at $P$ because this question depends on how I label $P$. Like I said above, there is no quantum mechanics here. Once we add quantum mechanics, Nature itself enforces invariance under diffeomorphisms by shrouding ill-formed questions with black holes.

What is quantum gravity about?

The moral is that, in quantum gravity, spacetime fluctuates. If I zoom in enough, these fluctuations are so violent they form black holes, so there are no local observables. This is also suggested by the classical invariance of general relativity under relabelling points. But if spacetime itself is fluctuating, what is quantum gravity even about? We’ll finish by describing a few possibilities.

  1. Black holes. Black holes are the guardians of the secrets of quantum gravity, appearing whenever we try to probe below the Planck scale. But they are also guides, providing generous (albeit indirect) clues such as the Bekenstein-Hawking entropy. Understanding how they store and release information is one of the keys to making progress in quantum gravity.
  2. Non-local observables. Even though local observables do not make sense when spacetime fluctuates, there are non-local observables that do. The basic idea is that certain spacetimes have regions which are so large or far away they are protected from fluctuation, since it would take too much energy to wobble them. Sitting in this non-fluctuating part, we can attach an electron-detector lure to a fishing rod, and dangle it near the point $P$. The lure will fluctuate in position, but because I sit somewhere stable, the whole procedure is well-defined. [These fishing-rod measurements are called dressed observables in the literature. There are other sorts of non-local observables, but I won’t discuss them here.]
  3. Emergent spacetime. I have somewhat understated the problem with microscopes. The formation of black holes is a picturesque way of saying that general relativity is inconsistent at high energies, and some new theory has to kick in at (or before) the Planck length. [See the appendix for more on this argument.] The most promising candidate is string theory, which makes the wackadoo prediction that spacetime is build out of tiny, vibrating strings. If they have a length $\ell_s \gtrsim \ell_P$, our microscopes will see them before we start making black holes. When spacetime materializes from some very different looking theory at high energies, we say it is emergent.
  4. Quantum shapes. In emergent theories, the macroscopic universe only appears as we zoom out from something fundamentally different. Alternatively, we can directly try to quantize the shape of spacetime, i.e. treat it as a fuzzy variable like the position of an electron. This is technically challenging since there many ways for space to look, but in loop quantum gravity, the challenge is surmounted by chunking spacetime into discrete, graph-like structures called spin networks, whose edges are around a Planck-length long. This is a very different way the theory can fail to make sense below the Planck scale. Look any closer, and you see nothing at all!

Appendix: QED, graviton exchange and the Planck energy

In this appendix, I’ll explain why combining gravity and quantum mechanics is not a problem until we hit the Planck scale. This is slightly more advanced, but we will still be outrageously heuristic. We start by discussing the quantum theory of electromagnetism, aka quantum electrodynamics (QED). We can then generalize to gravity, seeing what works and what doesn’t.

Quantum electrodynamics

Suppose two electrons pass by each other. Classically, they are repelled because each generates a field the other responds to. Quantum-mechanically, you cannot have action at a distance, and fields are replaced by messenger particles, that like carrier pigeons, alert the electrons to one another’s presence. For QED, the messenger particle is a photon, so the electrons throw some number of photons back and forth.

Without worrying too much about the details, let’s suppose the probability of exchanging a single photon is $\alpha$. We call $\alpha$ the coupling constant. What’s the probability of exchanging any number of photons? It’s just the sum of probabilities for the different number of exchanged photons:

\[\alpha + \alpha^2 + \alpha^3 + \cdots = \frac{\alpha}{1 - \alpha},\]

where we used the geometric series, assuming $\alpha < 1$. This is a very loose order-of-magnitude estimate for the probability $P$, and other constants can appear out the front, so $p \approx C\alpha/(1-\alpha)$, and this will be a reasonable probability with $p \leq 1$ for an appropriate $C$. But if $\alpha \to 1$, $p$ becomes infinite. No constant out the front can save us! When the probabilities add up to something sensible, we say the theory is unitary. When $\alpha = 1$, unitarity breaks down.

So, to check if QED is a sensible, unitary theory, we need to find $\alpha$. Recall Couloumb’s law for the electrostatic repulsion between two electrons:

\[F = \frac{k_ee^2}{r^2},\]

where $k_e= 9\times 10^{9} \text{ N}\cdot \text{m}^2/\text{C}^2$ is the Coulomb constant. In the quantum world, electrons do not have exact positions, and $r^2$ is not relevant to the probabilities. Instead, the probability of exchange is governed by the numerator, $k_e e^2$. [Technically, we usually take electrons to have perfectly well-defined momenta, $\Delta p= 0$, so by Heisenberg’s principle, $\Delta x = \infty$. They could be anywhere!]

It’s clear that $k_e e^2$ has dimensions. A quick calculation shows that

\[[k_e e^2] = [F r^2] = \left(\frac{ML^2}{T^2}\right) L,\]

where the term in brackets is the dimension of energy. The coupling constant $\alpha$ is a probability, i.e. a dimensionless number, so we need to make $k_ee^2$ into something dimensionless, and we can use the fundamental constants involved in the problem to make that happen. Since we’re doing quantum mechanics $\hbar$ is involved, and since photons are involved we can use the speed of light $c$, with dimensions

\[[\hbar] = \left(\frac{ML^2}{T^2}\right)T, \quad [c] = \frac{L}{T}.\]

We can combine these to form the dimensionless combination $k_ee^2/\hbar c$, since

\[\left[\frac{k_e e^2}{\hbar c}\right] = \left(\frac{ML^2}{T^2}\right) L \cdot \left(\frac{ML^2}{T^2}\right)^{-1} \frac{1}{T} \cdot \frac{T}{L} = 1.\]

The resulting coupling constant $\alpha$ is called the fine structure constant, and has a numerical value

\[\alpha = \frac{k_e e^2}{\hbar c} \approx \frac{1}{137}.\]

Since this is smaller than $1$, QED is unitary. Of course, we have dramatically simplified QED to make this argument work, but when we include all the technical details, the gist is the same: if the coupling constant $\alpha$ is too big, then adding up the probabilities for different processes to occur gives a nonsensical answer.

Quantum gravity

Quantum gravity works the same way, but instead of the photon, the messenger particle for gravity is the graviton. This theory is a bit more complicated than quantum electrodynamics, but uses the same underlying formalism. Thus, we can combine quantum mechanics and gravity into a theory of graviton exchange. This is a perfectly reasonable thing to do. The problem is that unitarity breaks down at the Planck scale.

The argument is very similar to QED. Two massive particles that pass each other exchange gravitons to signal their gravitational presence. If the probability of exchanging a single graviton is $g$, then the probability of exchanging any number is

\[P \propto g + g^2 + g^3 + \cdots = \frac{g}{1-g}.\]

Once again, unitarity will break down if $g$ approaches $1$. For particles of mass $M_1$ and $M_2$, Newton’s law of gravitation is

\[F = \frac{GM_1 M_2}{r^2}.\]

Repeating the argument from QED, we throw away the $r^2$ and divide by $\hbar c$, where the inclusion of $c$ in the dimensional analysis is justified because gravitons also travel at the speed of light. We discover that the coupling constant for quantum gravity is

\[g = \frac{GM_1M_2}{\hbar c}.\]

This looks very similar to the fine structure constant $\alpha$. If the masses are small, like the charges $e$, we should get a sensible answer right?

The subtlety lies in how we interpret the masses $M_1$ and $M_2$. If they are the rest masses, then for all known elementary particles this is indeed less than one. But gravity doesn’t distinguish between rest mass and the relativistic mass! We already saw this in action when we “derived” the radius of a black hole. So, if we collide two energetic particles, each with relativistic mass $M = E/c^2$, the probability of exchanging a single graviton is approximately

\[g \sim \frac{G E^2}{\hbar c^5}.\]

If we crank up the energies until $g \approx 1$, unitarity will break down. More precisely, this happens when

\[g \sim \frac{G E^2}{\hbar c^5} \sim 1 \quad \Longrightarrow \quad E \sim \sqrt{\frac{\hbar c^5}{G}} = E_P,\]

where we have defined the Planck energy $E_P = \hbar c/\ell_P$. This is the energy created by a microscope probing around the Planck length! Numerically, $E_P/c^2 \approx 20 \, \mu\text{g}$, around the mass of a flea egg as many sources inform me. So, although quantum gravity makes perfect sense at low energies — it is just graviton exchange — at the Planck scale, the theory breaks down and becomes non-unitary. This is where the mystery starts! The graviton picture also explains exactly what is creating the black holes when our microscope becomes too powerful. It is the quanta of gravity themselves.

Written on October 2, 2020
Physics   Hacks