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A Helmholtz coil is an arrangement of two circular coils that produces a magnetic field in the center which is locally uniform in direction and magnitude, or at least nearly so. The configuration is optimal when the radius of each coil is equal to the separation between coils.

If the coils were replaced with massive rings, would this also produce a locally uniform gravitational field in both direction and magnitude? Or would a different diameter to separation be better?

Is there a well-recognized name for this configuration of masses?

uhoh
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2 Answers2

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The charge moving in a circle produces a magnetic dipole, and it is the proximity of the two magnetic dipoles that produces an approximately constant magnetic field in between the two coils.

However a ring of matter does not produce a gravitational dipole, unsurprisingly since there is no such thing as a negative mass so the analogous gravitational dipole doesn't exist. Indeed at the point exactly between the two rings of matter the gravitational field would be zero since the gravitational attractions to the two rings would be equal and opposite.

The electromagnetic analogy would be to consider the electric field created by two charged rings. The geometry of this field would be the same the same as the geometry of the gravitational field created by two massive rings.

John Rennie
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  • Clearly I have not had enough coffee this afternoon! What brought this up was my discussion in the first paragraph of this question. I meant to ask more if there is an arrangement of masses that creates a more uniform field than a planet (effectively a point-source). Should I adjust the wording of the question? More importantly, is there a configuration of finite masses at finite distances to do this, even though it would not look like a pair of Helmholz-coils? There is some math here that's important. Perhaps I should refactor the question. – uhoh Aug 11 '17 at 06:59
  • I'm going to accept, and work on a separate question based in math, post-coffee. The answer is concise and the insight is helpful, so its better to keep as-is and ask again. – uhoh Aug 11 '17 at 07:09
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    @uhoh: the canonical uniform gravitational field is created by an infinite massive sheet – John Rennie Aug 11 '17 at 07:13
  • Thanks. I'm thinking in a practical way without infinities, rather than theoretical. I've just asked Is there any efficient additive way to improve field uniformity from a single charge sign? and I suspect there is a nice mathematical answer. – uhoh Aug 11 '17 at 07:42
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Fields scale as $M/r^2$, so scaling a structure's dimensions up by $\alpha$ and decreasing its mass density by $\alpha$ leaves the field unchanged. So the field at the center of a cylinder whose diameter increases linearly along its length (I initially incorrectly suggested exponential) and whose mass-density is decreasing inversely with length will have the same constant non-zero field everywhere along its length.
A linear change with length means, of course, that its a simple cone. We can expand the angle at the vertex until the cone becomes a sphere. So if you take a planet with a hole through the middle that happens to have a mass density inversely proportional to radius, then the field will be constant and reverse sign at the center. In fact, the hole doesn't have to go through the middle. The field everywhere has a constant magnitude, only its angle changes. Actually I read somewhere that the earth's density has a slight tendency towards a 1/r dependence and the field in the first 1000 miles or so doesn't change much, but I don't have a reference.
If we talk about a conical shell then the area mass density has to be constant for the constant field scaling to apply. So next time you eat an ice cream cone please be aware that its gravitational field along its axis is roughly constant.
Thinking of a simple structure more analagous to the Helmholz configuration, the field along the axis of a ring radius R goes as $GMx/(R^2+x^2)^{3/2}$. If we place the point mass, $m$, on the axis at distance, $d$, from the ring, the total field is $GMx/(R^2+x^2)^{3/2} + Gm/(x-d)^2$. We can now choose $m$ and $d$ and a position $p$ to set the first, second, and third derivatives to zero. I had to solve this numerically. The answer comes out to be m/M = 1.75620, d/R = -1.41456, p/R = 0.202622. The resulting field is 0.862275*GM/R^2

plot of normalized field vs. normalized position

Roger Wood
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    This is lovely, a donut plus its hole. I suppose if the point were instead a small ring one could call it a pair of "Helmholz coil-like mass configuration". – uhoh Oct 20 '20 at 07:59
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    @uhuh Yes, with slight readjustment it should work just as well with a solid donut. it's interesting that you can expand the point source into a sphere without changing its external field. And I guess you can do this for any object by expanding each element of mass into a corresponding sphere. So the line ring can be expanded into a solid donut wihout changing its field. However, this particular donut would end up with a somewhat higher mass density on its inner side. – Roger Wood Oct 24 '20 at 05:16
  • always thinking I see, thanks! ;-) fyi on the orbit question as far as I can tell, answers so far use theoretical 2D surface distributions of infinite 3D density. I'm thinking about asking for an example of a realistic 3D distribution of normal matter of variable but finite, and of uniform density. I'm still not convinced real-world examples exist. I'll ping you if/when it gets posted. – uhoh Oct 24 '20 at 05:54