In an answer to a question about the overall charge-neutrality of the universe, benrg writes,

A nonzero net charge density is incompatible with the cosmological principle. Unlike the gravitational field of a uniform mass distribution, which can be locally gauged away, the electromagnetic field of a uniform charge distribution has to increase linearly with distance from a privileged center point, and the field is locally measurable.

The reason for the homogeneous and isotropic initial conditions of the big bang isn't known, but at least this argument shows that whatever can produce those conditions will also produce a neutral universe.

From the similarity between the classical forms of Newton's gravitation and Coulomb's electrostatics — both of which are a coupling constant, a product of charges, and an inverse-squared distance — this distinction is not obvious. "Gauging away" a field has always been right at the cusp of my theoretical skill, and I'm out of practice. Can someone explain to me why these two fields behave differently?


2 Answers 2


The relevant difference between gravitational and electrostatic forces is that a gravitational field accelerates everything equally, whereas an electric field produces different accelerations on objects of different charge-to-mass ratios. This means that differently accelerated observers can disagree on the values of the gravitational field, but everyone must agree on the values of the electric field.

For a uniform mass density, every freely falling observer finds that they are at the center of the gravitational field.

To see this, let $\rho$ be the uniform mass density and consider a freely falling observer A at some position $\vec r_\mathrm A$. Assuming isotropic boundary conditions at infinity (in accordance with the cosmological principle), observer A would say that the Newtonian gravitational field at position $\vec r$ is $$\vec g_\mathrm{A}(\vec r) = -\frac{4\pi G\rho}{3}(\vec r-\vec r_\mathrm A).\tag{1}$$ Note that this expression correctly says that the gravitational field is $0$ at $\vec r_\mathrm A$: this is because observer A is freely falling, so from their point of view, local objects are unaccelerated.

Now consider a different freely falling observer B located at $\vec r=\vec r_\mathrm B$. To observer A, the gravitational field at $\vec r_\mathrm B$ is $$\vec g_\mathrm{A}(\vec r_\mathrm{B}) = -\frac{4\pi G\rho}{3}(\vec r_\mathrm{B}-\vec r_\mathrm A).\tag{2}$$ But to observer B, the gravitational field at any position $\vec r$ is equal to that measured by observer A minus observer B's own acceleration with respect to observer A. Due to free fall, observer B accelerates at precisely $\vec g_\mathrm{A}(\vec r_\mathrm{B})$ relative to observer A, so to observer B, the gravitational field is $$\vec g_\mathrm{B}(\vec r) = \vec g_\mathrm{A}(\vec r) - \vec g_\mathrm{A}(\vec r_\mathrm{B}) = -\frac{4\pi G\rho}{3}(\vec r-\vec r_\mathrm B).\tag{3}$$ Note that this expression similarly correctly says that the gravitational field is $0$ at $\vec r_\mathrm B$.

From equations (1) and (3), it is clear that every freely falling observer finds that they are at the "center" of the gravitational field. (This is analogous to how every observer in an expanding Universe finds that they are at the center of the expansion.) A uniform mass density is thus compatible with a homogeneous and isotropic universe.

For a uniform charge density, there is a privileged location that every observer finds to be the center of the electric field.

You can make an electrostatic analogue to equation (1). Some observer A at position $\vec r_\mathrm A$ would say that the electric field at position $\vec r$ is $$\vec E(\vec r) = \frac{\rho}{3\epsilon_0}(\vec r-\vec r_\mathrm A)\tag{4}$$ (again assuming isotropic boundary conditions), where $\rho$ is now the uniform charge density. The problem is that a freely falling, electrically neutral observer B at a different position $\vec r_\mathrm B$ would be unaccelerated with respect to A (neglecting gravity for simplicity), so they would agree on all accelerations measured by observer A, and hence would agree that the electric field is given by equation (4). Even if observer B were accelerated, they would agree with observer A on the differences in acceleration between particles of different charge-to-mass ratios, and so would still agree that the electric field is given by equation (4). Thus, every observer would say that the privileged position $\vec r_\mathrm{A}$ is the center of the electric field.

  • 1
    $\begingroup$ To summarize: The gravitational field does not have a local meaning, it is an artifact of your coordinate system. But the electromagnetic field does (it looks different in different reference frames but all frames agree if it is zero or non-zero). $\endgroup$ Commented Jun 15 at 19:34
  • $\begingroup$ But wouldn't one "redefine" the notion of neutral to mean $\rho$? $\endgroup$ Commented Jun 16 at 10:32

I don't agree with the assumption that the electric and gravitational fields must be proportional to ${\bf r} - {\bf r}_0$ for some central point ${\bf r}_0$.

As discussed at Gauss's law in a uniform charge distribution extending infinitely in all directions and Why isn't an infinite, flat, nonexpanding universe filled with a uniform matter distribution a solution to Einstein's equation?, the problem of determining the electric or gravitational field for a uniform source going out to infinity is simply underspecified; you need to specify boundary conditions to make the problem well-posed, and it isn't obvious to me that imposing spherical symmetry about some point is any more natural that any other choice of boundary condition. E.g. the solution ${\bf E} = \rho\, x\, \hat{\bf x}$ also solves Gauss's law, and it's fully translationally invariant in every direction except for the $x$ direction. So arguably it's even more "natural" than the spherically-symmetric boundary condition. (I'm not arguing that this boundary condition is actually more natural, but just that there isn't one single overwhelmingly natural choice.)

Relatedly, the Coulomb integral diverges for a uniform source field, which is another hint that the vector fields are not well-determined.

  • 3
    $\begingroup$ The $\vec f\propto\vec r$ behavior holds for isotropic boundary conditions at infinity, as I noted, but it's true that more general boundary conditions can produce different outcomes. However, note that your example still produces a privileged center -- the center is just a plane instead of a point. And if that were a gravitational field, it would still be "gauged out" as described in my answer, such that it would be consistent with homogeneity. $\endgroup$
    – Sten
    Commented Jun 14 at 22:24
  • 1
    $\begingroup$ Yes, the problem is ill-posed and I knew that when I wrote my answer (which I just rewrote). I was just trying to communicate the gist of the difference and I didn't realize it would attract the amount of attention it did. There must be a paper somewhere that makes a better version of the argument. $\endgroup$
    – benrg
    Commented Jun 14 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.