Wikipedia:Reference desk/Archives/Mathematics/2024 July 23

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July 23

Abel's reciprocity relation

Do we have an article that covers Abel's reciprocity relation for differentials of the third kind? Tito Omburo (talk) 18:37, 23 July 2024 (UTC)

Google Books and Google Scholar searches for "Abel's reciprocity relation" do not yield any results,^[1][2] so is this perhaps better known under a different name?  --Lambiam 21:22, 23 July 2024 (UTC)

I've seen it called the "first reciprocity law", but that's a bit non-specific. Tito Omburo (talk) 22:31, 23 July 2024 (UTC)

In this book I find a treatment of "The Reciprocity Theorem" for "Abelian differentials of the second and third kind", and in this one one of what is called "the reciprocity law for differentials of the first and third kinds". I'm only vaguely familiar with the first principles of differential geometry; the limited preview afforded by Google Books does not allow me to understand the notation and see how these two theorems, which superficially look rather different, are related. Both are apparently related to "Abel's theorem" on sums of integrals of certain types of function. We appear to treat only a very simple application of the latter theorem under the name "Abel's identity", so I am afraid the answer to your question is negative.  --Lambiam 09:29, 24 July 2024 (UTC)

Type of curvature?

Where $b'={\frac {a^{2}}{b}}$ (polar radius of curvature {RoC}) and $N(\phi )$ is the normal RoC, what is ${\frac {b'}{N(\phi )}}$ (especially the vertex latitude, ${\frac {b'}{N(\phi _{v})}}$)? Since it involves ${\frac {1}{N(\phi )}}$ it would suggest a type of curvature, rather than RoC—?
Likewise with geocentric latitude, $\psi$, there is ${\frac {b}{R(\psi )}}$ (with $R(\psi )$ being the geocentric radius).
This is based on $N(\phi )\cos(\phi )=R(\psi )\cos(\psi )$ (so, technically, isn't N the geographic prime horizontal RoC and R, besides being the geocentric radius, is also the geocentric prime horizontal RoC?). -- Kaimbridge (talk) 18:43, 23 July 2024 (UTC)

The ratio ${\frac {b'}{N(\phi )}}={\frac {a^{2}}{bN(\phi )}}$ can be rewritten as ${\sqrt {\frac {1-e^{2}\sin ^{2}\phi }{1-e^{2}}}},$ where $e={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$ is the eccentricity of the ellipsoid. Eccentricity is a dimensionless quantity, also when the radii $a$ and $b$ are expressed using units of length such as kilometres or miles, and so is this ratio. At the poles ($\phi =\pm {\frac {\pi }{2}}$) it equals $1$ and at the equator ($\phi =0$) it equals ${\frac {b}{a}}.$ If the radii are dimensioned lengths, then so is curvature; its dimension is then the inverse of length. Therefore there is no plausible way to interpret this dimensionless ratio as a form of curvature.  --Lambiam 21:05, 23 July 2024 (UTC)

But I thought the root definition of curvature is that its radius is its inverse, and anything else, here b', is just a modifier.

What about the idea of $N$ and $R$ being different prime horizontal RoC (in the case of the foundational, parametric $\beta$ latitude, the prime horizontal RoC is just $a$)? -- Kaimbridge (talk) 03:57, 24 July 2024 (UTC)

I do not understand what you are asking. A geometric interpretation of these quantities? Indeed, the radius of curvature is the inverse of the curvature; my point was that if one is dimensioned, so is the other, while the ratio about which the question appeared to be is inherently dimensionless, so it cannot be some type of curvature. I also do not understand what you mean by "just a modifier". It is the prime-vertical radius of curvature at the pole, which you proceed to divide by that at at geodetic latitude $\phi .$  --Lambiam 08:17, 24 July 2024 (UTC)

By "just a modifier", I mean ${\frac {1}{N(\phi )}}$ prime-horizontal curvature, so b' is a modifier of that curvature.

Okay, so what about $R(\psi )$ in this situation? Would $R(\psi )$ in ${\frac {b}{R(\psi )}}$ be considered the geocentric prime horizontal RoC?

In terms of purpose/context, ${\frac {b'}{N(\phi )}}$ and ${\frac {b}{R(\psi )}}$ are used (as vertex latitudes $\phi _{v}$ and $\psi _{v}$) in the geographic and geocentric (respectively) calculation of geodetic distance (rather than the usual $\beta$ based, parametric calculation, which has a neutral value, ${\frac {a}{a}}$), all using the same, iteratively found, geodetically adjusted longitude difference ($\Delta \Lambda =\Delta \lambda +{\mathcal {\ddot {A}}}$). -- Kaimbridge (talk) 06:54, 25 July 2024 (UTC)

A radius of curvature at a given spot is the radius of an osculating circle. It can be viewed as a vector from the centre of that circle to the given spot. If the direction of this RoC, viewed as a vector, is the vertical direction, it is a vertical RoC. (If, moreover, the plane of the osculating circle is perpendicular to the meridian through the given spot, so it intersects the ellipsoid along the east–west direction, it is the local prime-vertical RoC.) Unless the ellipsoid is a sphere, the geocentric radius at any other spot than the poles or equator, viewed as a vector from the centre to that spot, is not in the vertical direction, so it is not the radius of a locally osculating circle and it is not particularly meaningful to interpret it as either vertical or as a RoC, let alone both.  --Lambiam 10:41, 25 July 2024 (UTC)