This is inspired by 3-dim. Newtonian gravity, but I will stick to the 1-dim. case (so as not to overload the discussion with technical details; I also omit various constants).

The gravitational potential $\phi$ of an infinite, constant matter distribution $\rho = \text{const.}$ shall be derived using the Poisson equation

$$ \nabla^2 \phi = \rho $$

The general solution is

$$ \phi_a(x) = \frac{1}{2} (x-a)^2 + \phi_0 $$

where $a, \phi_0$ are arbitrary constants.

This breaks translational invariance and creates observable effects.

Using the harmonic oscillator potential and the resulting linear force we find the equation of motion for an observer in free fall

$$ \ddot{x}_a = \rho (x_a-a) $$

which is solved using $y = x_a - a$ and

$$ y(t) = A \cos(\omega t - \delta) $$

with $\omega^2 = \rho$ and arbitrary integration constants $A, \delta$.

For observers $n = 1, 2, \ldots $ we can define their orbits $y_n(t)$ and mutual distances

$$ d_{mn}(t) = y_m(t) - y_n(t) $$

Through mutual observations these observers can identify that there is a point around which their orbits oscillate.

This is what I call spontaneous symmetry breaking: for a translation invariant system, mathematics generates solutions - and with further arguments one can derive observable effects - that break this symmetry.

Because of translational symmetry, the coordinate $x=a$ is irrelevant; the position of the point is also irrelevant, since we can move it arbitrarily ($a \to a^\prime$ just creates new solutions); but we cannot get rid of the existence of a particular point around which the orbits oscillate.

We started with a translational invariant system, and breaking this invariance seems rather unphysical (for observers on elliptical Kepler orbits around a black star, this is somewhat different; they can also determine the position of the star, but here it is the star that breaks the invariance).

Restoring translational symmetry is possible: the linear force must vanish; therefore, the density must vanish, i.e. $ \rho = 0 $.

Modifying the system does not provide a nice solution, either. Going from the Poisson to the screened Poisson equation

$$ (\nabla^2 - \lambda) \, \phi = \rho $$

introduces a cosmological constant (sic!) $\lambda$ (which allows for a static 3-dim. universe filled with gravitating homogeneous dust; Einstein discussed that in his 1917 paper; others were also puzzled by artifacts in Newtonian spacetimes already in the 19th century).

This equation allows for a simple, translationally invariant solution

$$ \phi = -\frac{\rho}{\lambda} = \text{const.} $$

from which, however, the original case $\lambda \to 0$ cannot be recovered (for non-vanishing $\rho$).

Some thoughts:

  • Do you see any flaws or loopholes?
  • Are there more sophisticated methods to understand this system better?
  • What about this "anomaly" breaking a classical symmetry? Is it physically acceptable?
  • Are there more realistic systems where anything similar occurs? (in 3-dim. gravity it's similar but more involved)


  • 1
    $\begingroup$ Seems to me the potential in an infinite constant density distribution would have to be a constant due to symmetry. How could it depend on $x$? $\endgroup$
    – RC_23
    Commented Jul 27, 2023 at 14:01
  • $\begingroup$ Because that's the solution of the Poisson equation. $\endgroup$
    – TomS
    Commented Jul 27, 2023 at 14:05
  • $\begingroup$ Using appropriate boundary conditions or symmetry arguments, one could rule the $x^2$ potential out from the very beginning. But I don't want to do that b/c I am interested exactly in this artifact. The situation in 3 dim. is more involved (technically) but the core of the problem seems to be the same: constant homogenous mass density creates strange artifacts. $\endgroup$
    – TomS
    Commented Jul 27, 2023 at 14:21
  • $\begingroup$ If you don't impose boundary conditions then very similar ambiguities arise even if $\rho = 0$. In that case the solution to Laplace's equation is $\phi = a x + b$ and a test particle is pushed to the left or right depending on whether $a > 0$ or $a < 0$, breaking reflection symmetry. $\endgroup$ Commented Jul 27, 2023 at 15:21
  • 1
    $\begingroup$ Related question here. $\endgroup$
    – knzhou
    Commented Jul 27, 2023 at 18:13

4 Answers 4


This is an example of an ill-posed problem in mathematical physics (as defined by Hadamard). A "problem" in Hadamard's sense is a set of criteria (equations and possibly others) in which we provide an input and find a solution satisfying the criteria. For a problem to be "well-posed" it has to satisfy three criteria:

  1. A solution must exist.
  2. The solution must be unique.
  3. The solution must be stable, i.e., if we change the input by a "small amount" then the solution must change by a "small amount".

Any solution that is not "well-posed" is "ill-posed". So let's examine various versions of your problem and see if they're well-posed. (I'm going to ignore criterion 3 because it would require a more careful definition of "small changes"—some kind of well-defined metric on the input and solution spaces—and in the end criteria 1 & 2 will be the relevant ones for my discussion anyhow.)

Version 1:

Given a constant $\rho$, find $\phi: \mathbb{R} \to \mathbb{R}$ such that $\nabla^2 \phi = \rho$.

Here, $\rho$ is your input and $\phi$ is your desired solution. As you note, the general solution to this equation is $\phi = \frac{\rho}{2}(x - a)^2 + \phi_0$ for two arbitrary constants $a$ and $\phi_0$. This problem satisfies criterion 1 but fails criterion 2, and so is ill-posed.

Version 2:

Given a constant $\rho \neq 0$, find $\phi: \mathbb{R} \to \mathbb{R}$ such that $\nabla^2 \phi = \rho$ and which possesses translational symmetry.

The additional assumption seems reasonable, since $\rho$ also has translational symmetry. But as you correctly point out, no such solution exists, and so this problem fails to be well-posed by criterion 1.

Hadamard took the point of view that every mathematical problem corresponding to some physical or technological problem must be well-posed. If you take that view, the answer to your question "is this physically acceptable?" is "no."

That said, there are some problems, including many differential equation problems like yours, that intuitively seem to be physically reasonable but which are ill-posed by Hadamard's criteria. Physicists have often used so-called regularization methods to try to massage ill-posed problems into well-posed problems. (As a caveat, I must state that I'm not well-versed in these techniques. My sense is that these techniques are applied to problems where the solution is not stable, rather than non-existent or non-unique, but I don't know for sure. Hopefully I've given you enough to Google, at least.)


No anomaly, just a nice example of spontaneous symmetry breaking.

It is a prejudice that if a problem is symmetric under some group of transformations (here the translation group of the real line), then there must exist a solution which is invariant under that group of transformations.

When the problem admits solutions, the generic situation is that there is a family of solutions and that family is closed under the action of the group. This is a very general phenomenon both in mathematics and in physics, encompassing classical and quantum(-relativistic) systems.

That is called spontaneous breaking of symmetry.

The considered problem is of that type: you have a family of solutions and the group of translations sends each solution to another solution of the family.

If the family were made of a unique element, then it would be invariant under the group of symmetries.

But it is not the case here, as you did not give boundary conditions and the Poisson equation satisfies the existence and uniqueness problem only when boundary conditions are given. However, not every boundary conditions guarantee existence/uniqueness.

Looking at the family of solutions you found, to obtain a unique solution, you can try to impose that the solution is constant at $\pm\infty$.

These boundary conditions preserve translational invariace of the completed problem PDE + boundary conditions. Therefore the arising family of solutions should be translationally invariant as well, provided it is not empty.

As a matter of fact, if the two constants are different, then no solutions exist. In this sense the problem is ill posed as discussed in the other answer. However, if the constants coincide, then there is exactly one solution: the constant itself and this solution is translationally invariant as expected.

  • $\begingroup$ Oh, nice. So the differential equation doesn’t have a solution but a family of solutions. And the symmetry of the differential equation is shared by the family, not the individual solutions. Is that generally true? $\endgroup$
    – Dale
    Commented Jul 27, 2023 at 22:04
  • 2
    $\begingroup$ Yes, that is the most elementary, classical, case of spontaneous symmetry breaking. However if the solution is unique, it must be symmetric. The problem here is that the solution is not unique due to the absence of boundary conditions. $\endgroup$ Commented Jul 27, 2023 at 22:05
  • $\begingroup$ Thanks. I agree, it's spontaneous breaking of symmetry. $\endgroup$
    – TomS
    Commented Jul 28, 2023 at 6:31

I try to answer that here.

As explained by Valter the symmetry is broken at the level of an individual solution. Each solution has less symmetry than the original problem. A trivial example is Euclidean space where each solution

$$ \vec{x}(t) = \vec{x}_0 + \vec{v} \, t $$

breaks Galilei invariance, but Galilei transformations (translations, rotations, boosts) can be used to generate the entire family of solutions starting with one selected solution.

However, I see one key difference:

  1. In flat Euclidean space the problem is symmetric, the symmetry is broken by each solution for a test mass, the family can be reconstructed as group orbits of one solution
  2. In our case the problem is fully symmetric, but there are two steps:
  • First each potential with one central point $x=a$ breaks translational invariance.
  • Then the entire family of solutions for test masses based on one specific choice $\phi_{x=a}$ breaks translational invariance as well. Acting with a translation on one solution is forbidden, it does not generate a new solution for the same $\phi_{x=a}$.

I try to formalize that. $x$ is one specific solution, $X$ is the entire family of solutions.

In case (1) we have one solution $x$ and we generate the entire family $X$ as group orbit of $x$, here for Galilei transformations

$$ \text{Gal} \cdot x = X $$

In case (2) we have the family of solutions $X_a$ for $x=a$, and we generate the entire family $X$ acting on $\phi_a$ and $X_a$, now for translations

$$ \text{Trans} \cdot \phi_a = \Phi = \{\phi_{a}, a \in \mathbb{R} \} $$

$$ \text{Trans} \cdot (\phi_a, X_a) = ( \Phi, X) $$

In case of (2) it's not reasonable to act on $x_a$ alone because for a translation $T_d$ we have

$$ T_d \cdot X_a = X_{a+d} \neq X_a $$

therefore each family $X_a$ breaks translational invariance.

So we must act on both $\phi_a$ and $X_a$, that means we have something like

$$ (\Phi, X) = \bigcup_a \, (\phi_a, X_a) $$

Now, being one specific observer $x_{a, A_0, \delta_0}$, via physical observations and measurements I see (and all the others see as well) that my (our) entire $X_a$ breaks translational invariance (because everybody oscillates, nobody follows a straight line which would be the solution for a translational invariant problem).

And being a mathematician, I conclude that there are other families $X_{b \neq a}$ to which I as an observer in $X_a$ have no access.

We may compare that to Kepler orbits: As a rocket pilot I can change the orbit I am on, so I can explore different Kepler orbits around the sun. But I can't change the position of the sun. The difference to our case here is that the mass distribution of the sun explicitly breaks translational invariance of the original problem, whereas here symmetry breaking is an artifact when solving for $\phi$ based on a translational invariant mass distribution.


Through mutual observations these observers can identify the point around which their orbits oscillate.

I think they can't, actually. The trajectories can also be written

$$y_i(t) = \mathrm{Re}[A_i e^{\mathrm i ω t}]$$

By measuring relative positions (and time derivatives thereof), they can only determine $A_i-A_j$ for all $i,j$. In other words, they can determine the $A_i$ only up to an overall additive constant. Identifying the point $a$ is equivalent to identifying the trajectory with $A=0$, which they can't do.

This is similar to what happens in anti-de Sitter space in general relativity.

  • $\begingroup$ If they measure the mutual distances and exchange this information, they can at least see that for each reference observer $m$ the distances $d_{m, n \neq m}(t)$ follow a cosine law. That is sufficient to conclude that there is this kind of symmetry breaking; w/o any force they find straight lines instead of the cosine. $\endgroup$
    – TomS
    Commented Jul 28, 2023 at 5:47
  • $\begingroup$ I changed my post accordingly. $\endgroup$
    – TomS
    Commented Jul 28, 2023 at 7:20
  • $\begingroup$ @TomS I agree they can see that there is a force (they can measure $ρ$), but I don't think the symmetry is broken – again, this is analogous to anti-de Sitter space, where different timelike geodesics oscillate sinusoidally around the origin with different amplitudes, but they are all equivalent under the SO(3,2) symmetry and any of them can be taken as the origin around which the others oscillate. $\endgroup$
    – benrg
    Commented Jul 28, 2023 at 7:45

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