The Aharonov-Bohm effect concerns the electromagnetic fields, where a quantum particle can feel the effect of an electromagnetic field in regions where it vanishes, through the electromagnetic potential 4-vector $A_\mu$.

What is the analog of this in the case of general gauge fields? I would also like to know what is the analog of this for the case of the gravitational field (quantum particle in a classical gravitational field, governed by the Einstein equation).

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    $\begingroup$ The usual description of the A-B effect treats the EM field as a classical (non-quantum) field, which is often a good approximation. In the non-abelian case, that's not a good approximation at all, because quantum non-abelian gauge fields have very short range (we call this "confinement" or the "Higgs effect", depending on the details). But as a mathematical exercise, we can still generalize the usual description of A-B to classical non-abelian gauge fields. Are you asking for this mathematical generalization, even though it's not a good approximation physically? $\endgroup$ Commented May 9, 2020 at 23:53
  • $\begingroup$ @ChiralAnomaly yes, even mathematical generalizations would be helpful, although it seems important to point out why these generalizations are not good approximations physically in the answer. Thank you! $\endgroup$ Commented May 10, 2020 at 2:58

2 Answers 2


Review of the electromagnetic case

In the EM case, the Aharonov-Bohm effect can be deduced like this. The lagrangian for a non-relativistic charged particle is $$ L\sim \dot{\mathbf{x}}^2/2+\dot{\mathbf{x}}\cdot\mathbf{A}, \tag{1} $$ where the gauge field $\mathbf{A}$ is related to the magnetic field $\mathbf{B}$ by $$ \mathbf{B}=\nabla\times\mathbf{A}. \tag{2} $$ Using (1) in the path-integral formulation shows that the phase difference between two paths is sensitive to $\oint d\mathbf{x}\cdot\mathbf{A}$, where the integral is around the loop formed by the two paths. This shows up through the factor $$ \exp\left(i\oint d\mathbf{x}\cdot\mathbf{A}\right). \tag{3} $$ To reduce clutter, I absorbed a coefficient into the normalization of $\mathbf{A}$. The integral $\oint d\mathbf{x}\cdot\mathbf{A}$ is equal to the magnetic flux enclosed by the loop, so the phase of the complex number (3) can be non-zero even if it is invariant under local deformations of the loop — that is, even if the magnetic field is zero everywhere along the loop. This gives the Aharonov-Bohm effect.

Non-abelian generalization

The derivation reviewed above treats the EM field as a classical (non-quantum) field, which is often a good approximation. In the non-abelian case, it's not a good approximation. Even in the simplest model with just the gauge field and no matter ("Yang-Mills theory"), quantum effects lead to a spectrum of only massive particles ("glueballs") interacting with each other only through short-range forces, completely unlike the classical version of the model. The unfinished quest to understand why this happens is reviewed from one expert's perspective in The Confinement Problem in Lattice Gauge Theory.

Still, as a mathematical exercise, we can consider an analog of the Aharonov-Bohm effect using a classical non-abelian gauge field. In the non-abelian case, the quantity (3) is not gauge-invariant, because $\mathbf{A}$ transforms non-homogeneously under gauge transformations. A gauge-invariant generalization is the Wilson loop. To define it, take any closed path $C$ and subdivide it into infinitesimal segments $S$. The Wilson loop is$^\dagger$ $$ W(C) = \text{trace}\left( P\prod_S\exp\left(i\int_S d\mathbf{x}\cdot\mathbf{A}\right)\right), \tag{4} $$ where the trace is defined in a matrix representation and where the symbol $P$ means path-ordered: the factors in the product are multiplied in the same order as the segments around the path. The Wilson loop (4) is the non-abelian generalization of (3), and it has the same Aharonov-Bohm-like property: the phase of the complex number (4) can be non-zero even if it is invariant under local deformations of $C$.

$^\dagger$ This way of describing a Wilson loop looks unnatural because it tries to express elements of the Lie group in terms of elements $A_\mu$ of the Lie algebra. A more satisfying formulation describes the gauge field as a map (with special properties) from curves in spacetime to elements of the Lie group, and then $A_\mu$ and the covariant derivative both arise naturally by considering infinitesimal extensions or retractions of a curve at one of its endpoints.

Now, what is the non-abelian generalization of (1)? A non-abelian generalization of (1) is contrived in this paper, but I'll use a different approach. The model (1) is meant to be an approximation to quantum electrodynamics (QED). In QED, the interaction term has the form $$ \overline\psi\gamma^\mu A_\mu\psi = J^\mu A_\mu \tag{5} $$ where $\psi$ is a Dirac spinor's worth of fermion field operators, $\gamma^\mu$ are Dirac matrices, and $A_\mu$ is the gauge field. The combination $J^\mu=\overline\psi\gamma^\mu \psi$ is the electric current. This resembles the interaction term $\dot{\mathbf{x}}\cdot\mathbf{A}$ in the non-relativistic model (1). In the non-abelian generalization, $A_\mu$ is a matrix with components $(A_\mu)^{ab}$, and the spinor $\psi$ also carries a "color" index so that the interaction term is $$ \sum_{a,b}\overline\psi^a\gamma^\mu (A_\mu)^{ab}\psi^b = \sum_{a,b}(J^\mu)^{ab} (A_\mu)^{ab} \tag{6} $$ with current $(J^\mu)^{ab}=\overline\psi^a\gamma^\mu \psi^b$. The similarity between (5) and (6), combined with the constraint of gauge invariance, suggests the non-abelian case should exhibit an Aharonov-Bohm-type effect with the Wilson loop (4) in place of (3). That's not a proof, but the analogy is clear.

Gravitational analog

An online search for "gravitational Aharonov Bohm" finds several papers, including this proposal for an experiment. However, the analogies are not perfect. In perturbation-theory language, the gravitational field has spin 2, whereas the gauge fields considered above have spin 1, and that difference makes any analogy imperfect. I'll highlight one imperfect analogy.

What should we use as the gravitational analog of a region where the magnetic field is zero? One analogy could be a region where spacetime is flat — that is, where the Riemann curvature tensor is zero. But what arrangement of matter could possibly create a situation where the curvature of spacetime is non-zero inside a bounded region and zero outside?

It's surprisingly easy, if we use the same idealization that we implicitly use in the EM case. Recall that in the EM case, in order to make the field outside the solenoid strictly zero, the solenoid needs to be infinitely long. Mathematically, if the solenoid is infinitely long, we might as well discard that dimension of space and work in 2d space. The analog of this in general relativity is to work in 2+1 dimensional spacetime, and this version of GR has a remarkable property: any empty region of spacetime is always automatically flat! (I'm assuming a zero cosmological constant.)

More precisely, empty space is locally flat, but the circumference-to-radius ratio of a loop can still be affected by any matter that it encircles. This is easiest to visualize when the central mass is concentrated at a point, which gives a conical singularity. In this case, the 2d space can be visualized as a cone. The surface of the cone has zero intrinsic curvature (because we can construct a cone from a flat piece of paper without wrinkling it), but a circle centered on the apex has a circumference-to-radius ratio less than $2\pi$. The ratio depends on the amount of mass concentrated at the center.

That sounds similar to the Aharonov-Bohm situation, but a non-rotating point mass is the gravitational analoug of a non-rotating point charge in electrodynamics. We want a gravitational analog of a magnetic field, not of an electric field. Since a rotating charge makes a magnetic field, we can try to improve the analogy by considering a rotating mass. But then something strange happens: the spacetime outside a pointlike rotating mass admits closed timelike curves! For more about this, see Gravitation vs. Rotation in 2+1 Dimensions.

Other analogies are considered in knzhou's answer.


Since you've already received a good answer to the full question, I'll just add a bit more detail on the analogue of the Aharanov-Bohm effect for gravity. As noted in the existing answer and elsewhere, any analogy between gravity and gauge theory can't be perfect because the graviton has spin $2$.

On the other hand, we don't need a perfect analogy if we just want an analogue of the Aharanov-Bohm effect, because the standard setups for it (and its relatives, such as the Aharanov-Casher effect) all involve experimental equipment that breaks Lorentz invariance. After all, you would never analyze this effect in a frame where the solenoid was moving! If we don't bother to maintain manifest Lorentz invariance, we are free to butcher gravity until it takes the shape we want.

First analogy: Newtonian limit

In the standard Newtonian limit of general relativity, the metric reduces to $$ds^2 = - (1 - 2 \Phi) \, dt^2 + (1 + 2 \Phi) \delta_{ij} \, dx^i dx^j$$ where $\Phi$ is the Newtonian potential, and is sourced by matter in the standard way, $$\Phi(x) = \int d \mathbf{x}' \, \frac{\rho(\mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$ The action for a particle is $$S = - m \int \sqrt{-g_{\mu\nu} u^\mu u^\nu} \, d\tau$$ where $u^\mu$ is the four-velocity. In the nonrelativistic limit $u^\mu \approx (1, \mathbf{v})$, this becomes $$S \approx - m \int \sqrt{1 - 2 \Phi - (1 + 2 \Phi) v^2} \, dt \approx \int \frac{mv^2}{2} (1 + 2 \Phi) - m(1+\Phi) \, dt.$$ Neglecting the $\Phi v^2$ term as small, since both $\Phi$ and $v^2$ are, this is a standard "kinetic minus potential" Lagrangian, where the mass-energy $m$ and the gravitational potential energy $m \Phi$ are counted in the potential. Therefore, an object picks up an extra gravitational phase shift of $$\Delta \phi = - \frac{m}{\hbar} \int \Phi \, dt.$$ This phase shift is commonly measured by atom interferometry.

Is this the gravitational Aharanov-Bohm effect? According to this paper, it is. After all, it's a phase that can be measured by interferometry, which accumulates even when the gravitational field at the particles themselves vanishes. On the other hand, the phase shift doesn't have the $\mathbf{A} \cdot \mathbf{v}$ form, so it depends on the rate at which we traverse a path, while the usual Aharanov-Bohm effect doesn't. It doesn't have the same "geometric" flavor.

Second analogy: Gravitoelectromagnetism

To get something closer, we can back up and do a less crude approximation. In the weak field approximation to general relativity, where $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, the field $h_{\mu\nu}$ is sourced by matter as $$\bar{h}_{\mu\nu}(\mathbf{x}) = 4 \int d\mathbf{x}' \, \frac{T_{\mu\nu}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$ where we have defined the trace reversal $\bar{h}_{\mu\nu} = h_{\mu\nu} - \eta_{\mu\nu} h / 2$ and set $c = G = 1$. Now, this looks quite similar to how the vector potential is sourced by a current, $$A_\mu(\mathbf{x}) = \int d\mathbf{x}' \, \frac{J_{\mu}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}$$ but we have an extra tensor index. Now we can abandon Lorentz invariance to go further. Note that for nonrelativistic source matter with typical speed $u$, the stress energy tensor components are $$T_{00} \sim O(1), \quad T_{0i} \sim T_{i0} \sim O(u), \quad T_{ij} \sim O(u^2).$$ So in the nonrelativistic limit, we could choose to neglect the $T_{ij}$ (and hence the $\bar{h}_{ij}$) entirely. Then, since the stress-energy tensor is symmetric, $T_{i0}$ is redundant with $T_{0i}$, so we only need to keep track of the elements $T_{0\mu}$. This is the same number of degrees of freedom as a four-vector.

By expanding $\bar{h}_{ij}$ to leading order in $u$, we get the slightly more general metric $$ds^2 = - (1 - 2 \Phi) \, dt^2 + 2 (\mathbf{A} \cdot d \mathbf{x}) \, dt + ( 1 + 2 \Phi) \, \delta_{ij} dx^i dx^j$$ where $A_\mu = (\Phi, \mathbf{A})$ is the gravitoelectromagnetic (GEM) four-potential, and $$A_0 = \Phi = \int d\mathbf{x}' \, \frac{T_{00}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}, \quad A_i = \int d\mathbf{x}' \, \frac{T_{0i}(t_{\text{ret}}, \mathbf{x}')}{|\mathbf{x} - \mathbf{x}'|}.$$ That is, $T_{0\mu}$ sources $A_\mu$ just like $J_\mu$ sources $A_\mu$ in electromagnetism. Of course, the analogy is not perfect because $T_{0\mu}$ isn't a four-vector, so our $A_\mu$ doesn't have nice Lorentz transformation properties, but nonrelativistic source matter already picks a preferred frame anyway.

For a slowly moving test particle, a similar analysis to the Newtonian limit gives $$S \approx \int \frac{mv^2}{2} - m(1+\Phi) + m \mathbf{v} \cdot \mathbf{A} \, dt$$ where we are working to quadratic order in $u$ and the speed $v$ of the test particle. The coupling to $\mathbf{A}$ looks just like the coupling to the magnetic vector potential, which implies that the force on the particle obeys the usual Lorentz force law, but with charge $m$ and "gravito-electric" and "gravito-magnetic" fields, which are defined in terms of $\mathbf{A}$ just as in electromagnetism. For more on this idea, which is used to analyze precision relativity experiments such as Gravity Probe B, see here.

Finally, the phase shift due to encircling a gravito-magnetic flux is $$\Delta \phi = \frac{m}{\hbar} \int \mathbf{v} \cdot \mathbf{A} \, dt = \frac{m}{\hbar} \int \mathbf{A} \cdot d \mathbf{x} = \frac{m \Phi_B}{\hbar}.$$ Since $\mathbf{E}$ and $\mathbf{B}$ are defined analogously to electromagnetism, $\mathbf{B}$ obeys the Biot-Savart law with current density $T_{0i}$, which is the momentum density. That is, the gravitational analogue of a solenoid is a rotating cylinder. In a more geometric language, where we hadn't sliced away the Lorentz invariance, this would be described as a "frame dragging" effect. (Explicitly, the magnetic flux through a cylindrical solenoid is $\mu_0 A J$ where $A$ is the area and $J$ is the current density. The computation of the gravitomagnetic flux is identical, except that $J$ is replaced with $\sigma v$, the product of the surface mass density and velocity of the cylinder.)

So, is this the gravitational Aharanov-Bohm effect? Again, the analogy is decent, but not perfect. In the nonrelativistic limit, the phase shift indeed looks identical to that of the ordinary Aharanov-Bohm effect. And the gravitomagnetic field does vanish everywhere outside the cylinder. But there's one catch: the gravitoelectric field (i.e. the ordinary gravitational field) doesn't vanish outside the cylinder, while the electric field outside a solenoid does. This deficiency is because there are no negative gravitational charges, so our gravitational solenoid is like a regular solenoid if the electrons were rotating without any compensating protons. As with everything else, it can ultimately be traced back to the fact that the graviton has spin $2$ instead of $1$, since even-spin mediated forces are universally attractive. (For what it's worth, this ultimately comes back to bite electromagnetism as well: the fact that charges can cancel out, leaving only currents, implies that it has two distinct nonrelativistic limits.)

In summary, while there are no perfect analogies to the Aharanov-Bohm effect, there exist at least two approximate ones. The first is measured in real experiments; the second arises from a formalism used to analyze other real experiments. Would it someday be possible to measure the gravitomagnetic phase shift through atom interferometry? It seems very difficult now, since it is penalized by a factor $u$ of the source matter speed, but who knows in the future!

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    $\begingroup$ Thanks for this detailed answer! The analogy between the gravimagnetic and ordinary magnetic field holds in the nonrelativistic limit, where some components of $g_{i j}$ and $T_{i j}$ can be neglected in comparison with others. I wonder, whether there is some way to couple $h_{\mu \nu}$ to some higher-dimensional object, that a worldline? $g_{\mu \nu}$ is a symmetric tensor, not a 2-form, so the guess with worldsheet seems not to be correct, but maybe there is something similar? $\endgroup$ Commented May 16, 2020 at 9:35
  • $\begingroup$ @spiridon_the_sun_rotator Certainly. The generalization to an $n$-dimensional object comes from including $\sqrt{- \text{det} \, g}$ in the worldsheet integral, in order to get the worldvolume. (Just like how for a $1d$ world line we use $g_{\mu\nu}$ to get the proper length.) For weak gravity, one could then expand the determinant in $h$. However, in order to get something that looks like the Aharanov-Bohm phase shift, I think you still need to take the nonrelativistic limit. $\endgroup$
    – knzhou
    Commented May 17, 2020 at 18:50
  • $\begingroup$ A quick question: you are saying that if, in the second approach, we didn't slice away Lorentz invariance, the equation for $$\Delta\phi$ would still hold? Or does it only hold when we lose Lorentz invariance? $\endgroup$ Commented May 19, 2020 at 19:34
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    $\begingroup$ @TheQuantumMan Everything in the second approach breaks manifest Lorentz invariance, because it all comes from the GEM four-potential $A_\mu$, which is not a four-vector. It doesn't transform like the ordinary electromagnetic four-potential, so all the following results must be regarded as only valid in the frame of the source matter. $\endgroup$
    – knzhou
    Commented May 19, 2020 at 19:37

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