On a Riemannian manifold, a coordinate system is called "isothermal" if the Riemannian metric in those coordinates is conformal to the Euclidean metric:

$$g_{ij} = e^{f} \delta_{ij}$$

My question is: Why are such coordinate systems called "isothermal"? It must have something to do with classical thermal physics. I tried looking for a reason online, with no success.

It is well known that when the dimension $n=2$, there always exist isothermal coordinates, and this is probably where they were first introduced. So maybe the nomenclature has something to do with heat diffusion in the plane?

(The reason I ask is because I am planning to give a seminar talk next week giving a proof that such coordinates exist when $n=2$, and thought it would be nice to explain to the students where the name comes from...)

  • $\begingroup$ I keep trying to fix the display: the comma after the n=2 is too close. It looks fine in the preview. Any suggestions? $\endgroup$ – Spiro Karigiannis Jul 16 '10 at 13:47
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    $\begingroup$ As the display depends highly on which browser you are using and which computer you are using and which set of fonts your are using, I think trying to prettify the display is a futile exercise. Just make sure it is syntactically correct. $\endgroup$ – Willie Wong Jul 16 '10 at 14:00

Isothermal coordinates are harmonic. In other words, it solves $\triangle_g u = 0$. So locally it is a stationary solution of the heat equation. In physics, for a steady state distribution of temperatures, each level set is called an isotherm.

  • $\begingroup$ Thanks! I should have thought of that, since I knew that isothermal coordinates are harmonic. In general, though, one can consider "harmonic coordinates" which are not necessarily isothermal. So it may not be the best nomenclature, but I guess it's here to stay. $\endgroup$ – Spiro Karigiannis Jul 16 '10 at 14:50
  • $\begingroup$ Well, in two dimensions where the original studies were done, there's no obstruction to conformally flat, so harmonic and isothermal coordinates are more or less the same (one also needs the two coordinates to be harmonic conjugates of each other). Also now isothermal coordinates also picked up the additional meaning to be coordinates in which it is conformally Euclidean. So it does mean something more specific than harmonic. $\endgroup$ – Willie Wong Jul 16 '10 at 15:14

According to Gray, Abbena and Salamon, that's the name given to such coordinate systems by Gabriel Lamé in his 1833 study of heat transfer. The reason is, if you've got a thermally isolated surface of constant heat conduction, the constant coordinate lines are isotherms iff the coordinates are isothermal.


Isothermal coordinates on a two-dimensional manifold are not harmonic unless the manifold has zero mean curvature (as an immersed submanifold of Euclidean space, for example.) The plane is a minimal surface (well, you COULD screw around with the parametrization :D ) and so isothermal coordinates would be conformal or anti-conformal there.

In higher dimensions, isothermal coordinates may not exist but harmonic coordinates do; Einstein introduced them in (about) 1916.

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    $\begingroup$ Zero mean curvature makes no sense unless the manifold is immersed inside another Riemannian manifold, which one does not need to assume to define isothermal coordinates. It is true that if a surface is immersed in the Euclidean $\mathbb R^3$, with the induced metric, then in isothermal coordinates, the mean curvature vanishes if and only if the coordinate functions are harmonic. However, on any Riemannian manifold, one has a notion of a Laplacian, and it's true that locally one can always find coordinates such that the coordinate functions are harmonic functions. $\endgroup$ – Spiro Karigiannis Aug 8 '10 at 21:04
  • $\begingroup$ We all know that mean curvature is extrinsic rather than intrinsic. I believe I did say "as an immersed submanifold of Euclidean space" in my answer. DeTurck & Kazdan showed in 1981 that harmonic coordinates yield the best regularity properties and Jost & Karcher had some nice results. $\endgroup$ – Some Math Guy Sep 6 '10 at 4:14
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    $\begingroup$ Your answer still does not make sense. Given a Riemannian manifold, one can define isothermal coordinates and ask whether or not those coordinate functions are harmonic (with respect to the Laplacian of the Riemannian metric) without needing to embed it into a Euclidean space, so there is no notion of mean curvature. A $2$-manifold with a metric can have isothermal coordinates and be isometrically immersed into $\mathbb R^3$ in two different ways, one which has mean curvature zero and one which does not. $\endgroup$ – Spiro Karigiannis Apr 1 '11 at 14:10

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