Gauss says that there should be no field within a hollow shell, spherical or cylindrical. But when I integrate the contributions to gravitational potential from a cylindrical shell to an off-axis point within the cylinder, I get an integral that looks like this:$$ \phi ~~= ~~ \int^{\pi}_{0} {\dfrac{\mathrm{d}x}{\sqrt{\alpha ^{2}+\alpha }~-~2\alpha \cos {\left( x \right)}}} \,,$$ where $\alpha$ is the ratio between the radius of the off-axis point to the cylinder radius.

The fact that this integral produces different potentials for different values of $\alpha$ means that there is a potential gradient within the cylinder, and that a particle will feel a gravitational force.

Where did Gauss (or maybe me) go wrong?

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    – Yashas
    Commented Mar 4, 2017 at 5:21
  • $\begingroup$ Is it a finite cylinder? You know that if you calculate for a ring of material, you get zero for everywhere inside the ring in the plane of the ring. A cylinder is a bunch of rings and rings that are not on the plane of interest can be thought of as pairs symmetrically on either side of your point of interest and their contributions will all sum to zero. I would guess you have set up the problem wrong or missed a symmetry. $\endgroup$ Commented Mar 4, 2017 at 8:18

2 Answers 2


This mixes up several distinct ideas, but makes a good conceptual or educational question. Firstly, "Newton's shell theorem" says the gravitational field inside a spherically symmetric distribution of mass is zero. This will not be true for general shapes; in particular intuitively I wouldn't expect it for a cylinder, apart from at the very centre. So where is the misconception?

Gauss's law for gravity (integral version) says$$ \oint_{{\partial}V} \vec{g} \cdot \mathrm{d} \vec{A} ~~=~~-4\pi GM \,.$$ In words, the total flux of the gravitational field across a closed surface equals (up to a constant) the enclosed mass. So if we take a surface inside the cylinder, we know the enclosed mass is $M=0$, so the law tells us the total flux of the gravitational field across this surface is zero, but not that the field itself is zero.

  • 1
    $\begingroup$ If the cylindrical shell is infinitely long with uniform mass density, then the interior field is also identically zero. Perhaps this is what OP meant. $\endgroup$
    – J. Murray
    Commented Feb 6, 2018 at 1:45

The integral in your question looks like the potential within a ring of matter,

$$ \phi(\alpha) = -\frac{G M}{\pi R} \int_0^\pi \mathrm{d}x \frac{1}{\sqrt{1 + \alpha^2 + 2 \alpha \cos(x)}}. $$

($\alpha$: distance from center relative to radius; $R$: radius; $M$: total mass of ring)

Inside a ring the gravitational field is indeed not zero (except for the center). Gauss’s law still applies, but as Colin MacLaurin pointed out there will be parts of the integration surface where the field points outwards and parts where it points inwards, which cancel each other. Here is a qualitative picture of the fieldlines in a plane perpendicular to the ring:

fieldlines of a ring mass, qualitatively

To obtain the potential within a cylinder, you must still integrate over the direction along its axis (taken as $z$ coordinate here):

$$ \phi(\alpha, z) = -2 G R \sigma \cdot \int_{z_0}^{z_1} \mathrm{d}z' \int_0^\pi \mathrm{d}x \frac{1}{\sqrt{R^2 \left[ 1 + \alpha^2 + 2 \alpha \cos(x) \right] + (z-z')^2}}. $$

($z$: coordinate along cylinder axis; $z_0$, $z_1$: bottom and top of cylinder; $\sigma$: mass per surface area)

If the cylinder is short, the pattern of fieldlines looks similar to the ring case. However, the center region where the field is weak becomes larger. In a cylinder of infinite length the field inside actually vanishes and the potential is constant. (Though the above integral won't give you any valid potential at all, because it diverges for $z_0 \to -\infty$, $z_1 \to +\infty$. You'd have to modify it a bit.)

In the case of infinite length you can in fact use Gauss's law to show that the field vanishes. For the integration surface, choose a smaller concentric cylinder. Then, for symmetry reasons, the field must have the same strength everywhere on it and either point radially inwards or outwards everywhere. No cancellations can happen. And since there is no mass inside the cylinder, you can conclude that the total flux as well as the field itself must be zero.


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