Surface with Ricci scalar equal to two

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Discussion Overview

The discussion revolves around the properties of a two-dimensional Riemannian manifold with a specific metric and its Ricci scalar. Participants explore the implications of the Ricci scalar being equal to 2 and whether this indicates that the surface in question is a sphere of radius r=1. The conversation includes mathematical reasoning, exploration of curvature, and the application of differential geometry concepts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the Ricci scalar as R = -1/r * d(e^-f)/dr and questions if a specific function leads to a surface that is a sphere of radius r=1.
  • Another participant suggests that the thread may be more appropriate for "Tensor Analysis & Differential Geometry," emphasizing the mathematical nature of the discussion.
  • A participant describes the embedding of the manifold as a surface of revolution in E^3 and discusses the implications for curvature and the Ricci scalar.
  • One participant asserts that the Ricci scalar for a sphere is 2/a^2, introducing a distinction between variable r and constant a.
  • Another participant presents a different expression for the Ricci scalar, indicating a potential disagreement on the calculations involved.
  • Further clarification is provided regarding the relationship between the Ricci scalar and Gaussian curvature, with references to specific results from differential geometry literature.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the surface and the calculations of the Ricci scalar. There is no consensus on whether the surface is definitively a sphere or on the correct expression for the Ricci scalar, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants reference various mathematical concepts and results from differential geometry, but there are unresolved aspects regarding the assumptions and definitions used in their calculations. The discussion includes multiple interpretations of the Ricci scalar and its implications for curvature.

Giammy85
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A two-dimensional Rienmannian manifold has a metric given by
ds^2=e^f dr^2 + r^2 dTHETA^2
where f=f(r) is a function of the coordinate r

Eventually I calculated that Ricci scalar is R=-1/r* d(e^-f)/dr


if e^-f=1-r^2 what is this surface?


In this case R comes to be equal to 2
I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2

So, is this surface a sphere of radius r=1?
 
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Cartan's method of curvature forms

Doesn't this thread belong in "Tensor Analysis & Differential Geometry"? It seems to involve a question about the mathematical theory Riemannian two-manifolds, not relativistic physics.

[EDIT: thanks to the unsung admin who moved this thread!]

(I corrected the formatting and made a small change in notation.)

Giammy85 said:
A two-dimensional Rienmannian manifold has a metric given by
<br /> ds^2= \exp(2f) \, dr^2 + r^2 \, d\theta^2<br />
where f is a function of the coordinate r.

Which can be embedded as a surface of revolution in E^3 as you probably know, at least locally (depending on f). To wit:
<br /> \left[ \begin{array}{c} h(r) \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]<br />
leads to the line element
<br /> ds^2 = \left(1 + {h^\prime}^2 \right) \, dr^2 + r^2 \, d\phi^2<br />
from which we obtain the ODE 1 + {h^\prime}^2 = \exp(2f) which you can solve to obtain h in terms of f.

Note that the euclidean formula for the circumference of the circle r=r_0 holds good, but (for reasons easy to understand from the embedding in case of a typical function h!) the usual formula relating dr to radial distance does not hold true, in general. In higher dimensions, BTW, in the context of gravitation physics, it is customary to refer to a radial coordinate with these properties as a Schwarzschild radial coordinate.

Giammy85 said:
Eventually I calculated that the Ricci scalar is...

This isn't homework, is it? Or even worse, a take-home exam problem? :frown: If so, you should have posted in the Homework Help forum, which has special rules!

The sphere is a space of constant Gaussian curvature (see the component R_{1212} of the Riemann tensor), which is an invariant property. See almost any differential geometry textbook for more information, e.g. Struik, Lectures on Classical Differential Geometry.

Here's the slick way to compute the curvature tensor (especially efficient in higher dimensions): from the line element
<br /> ds^2 = \exp(2f) \, dr^2 + r^2 \, d\phi^2, \; 0 &lt; r &lt; \infty, \; -\pi &lt; \phi &lt; \pi<br />
(for convenience I made a slight change in notation, and where the ranges of the coordinates are the maximal permissible) we can read off the coframe field
<br /> \sigma^1 = \exp(f) \, dr, \; \sigma^2 = r \, d\phi<br />
so that the line element is simply
<br /> ds^2 = \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2<br />
Taking the exterior derivatives of these one forms gives
<br /> d\sigma^1 = 0, \; d\sigma^2 = dr \wedge \phi<br />
But Cartan's first structural equations tell us that
<br /> d\sigma^j = -{\omega^j}_k \wedge \sigma^k, \;<br /> {\omega^j}_k = -{\omega^k}_j<br />
(since the exponential of anti-symmetric operator is a rotation). Plugging in the result we just obtained gives
<br /> -d\phi \wedge dr = dr \wedge d\phi <br /> = -{\omega^2}_1 \wedge \sigma^1 <br /> = -{\omega^2}_1} \wedge \exp(f) \, dr<br />
from which the (so(2) valued) connection one-form is
<br /> {\omega^1}_2 = -\exp(-f) \, d\phi<br />
(As an easy check, the exterior product of this with \sigma^2 = r \, d\phi does indeed vanish, as it should.) Taking the exterior derivative of this one-form gives
<br /> d{\omega^1}_2 <br /> = f^\prime \, \exp(-f) \, dr \wedge d\phi <br /> = \frac{f^\prime}{r} \, \exp(-2f) \, \sigma^1 \wedge \sigma^2<br />
But Cartan's second structural equation tells us that the (so(2) valued) curvature-two form is
<br /> {\Omega^j}_k = d{\omega^j}_k + {\omega^j}_m \wedge {\omega^m}_k<br />
where the second term drops out in two dimensions. We can now read off the Riemann curvature tensor from
<br /> {\Omega^j}_k = {R^j}_{kmn} \, d\sigma^m \wedge d\sigma^n <br />
Finally, lowering an index gives
<br /> R_{1212} = \frac{f^\prime}{r} \, \exp(-2f)<br />
which is the only algebraically independent component in two dimensions. Note that in the coordinate cobasis dr, \; d\phi this becomes R_{r \phi r \phi} = r \, f^\prime.

This method is called the "method of Pfaffians" in Struik, and the "method of one-forms" in MTW, and was introduced by Elie Cartan c. 1905, who later championed its use in gravitation physics. In the case of surfaces embedded in E^3, it should always give the same results as the well known theory developed by Gauss c.1820 but not published until 1827-8. See Flanders, Differential Forms with Applications to the Physical Sciences, Dover reprint, for details.

(Hmm... I see that the current version at the time of my post of the WP article on Gauss is incorrect: Gauss presented his famous memoir to the Royal Society of Gottingen in 1827, and it was published in their journal the following year; this paper predates the publications of both Lobachevski and Bolyai and greatly generalizes their work. A English translation with commentary can be found in Karl Friedrich Gauss, General Investigations of Curved Surfaces, Dover reprint. See also the superb comments in Spivak, Comprehensive Introduction to Differential Geometry. Hmm... actually the WP article is pretty damn bad overall IMO, even ignoring factual errors.)

Giammy85 said:
In this case R comes to be equal to 2
I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2

Unless stated otherwise, I always evaluate tensor indices wrt a frame field or coframe field (aka ONB of Pfaffians or one-forms or covectors, in order of increasingly modern terminology).

The sphere of radius a has Gaussian curvature R_{1212} = 1/a^2 (units of reciprocal area; components expanded wrt a coframe field). Solving
<br /> \frac{f^\prime}{r} \, \exp(-2f) = \frac{1}{a^2}<br />
gives
<br /> \sigma^1 = \frac{a \, dr}{\sqrt{a^2-r^2}}, \; 0 &lt; r &lt; a<br />
which agrees with line element
<br /> ds^2 = \frac{a^2 dr^2}{a^2-r^2} + r^2 \, d\phi^2, \; 0 &lt; r &lt; a, \; -\pi &lt; \phi &lt; \pi<br />
which we obtain from the obvious embedding in E^3
<br /> \left[ \begin{array}{c} \sqrt{a-r^2} \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]<br />
of the top half of a round sphere of radius a.

A Riemannian two-manifold of non-constant curvature cannot be any of S^2, \; E^2, \; H^2, which have curvatures of form 1/a^2, 0, -1/a^2 respectively (in any chart), where a&gt;0 is the "radius", in the first and third cases.
 
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mhm... I think it is a sphere of radius r=1
 
In the case of Riemannian two-manifold, the Ricci scalar is twice the Gaussian curvature, which makes 2/a^2 for a sphere of radius a. In my post above, r is a variable, the Schwarzschild radial coordinate on a surface of revolution, whereas a is a positive real constant.

Seriously, this wasn't a take-home problem exam, was it? (A possibility which unfortunately didn't occur to me until after I posted my detailed Post #2.)
 
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I get a different result for the Ricci scalar, viz

\frac{\partial rf}{re^{f}} where \partial r is differentiation wrt r.
 
Hi, Mentz114,

Mentz114 said:
I get a different result for the Ricci scalar, viz
<br /> \frac{\partial rf}{re^{f}}<br /> where \partial r is differentiation wrt r.

The Ricci scalar is R = \frac{2 f^\prime}{r} \, \exp(-2f) in the notation I used above. (The Ricci tensor, with components evaluated wrt the coframe field I gave, is diagonal with both diagonal components equal to the Riemann curvature component I computed.)

The result of Problem 3.4-19 in Struik implies that the Gaussian curvature is
<br /> K = \frac{R_{r \phi r \phi}}{\exp(2f) \, r^2} <br /> = \frac{r \, f^\prime}{ r^2 \, \exp(2f)}<br /> = \frac{f^\prime}{r} \, \exp(-2f) = R_{1212}<br />
as I claimed.

Exercise: write an orthogonal chart for the general Riemannian two-manifold in the form ds^2 = A^2 \, du^2 + B^2 \, dv^2, where A,B can be functions of u,v (although this isn't necessary; without loss of generality we could impose further restrictions), and adopt the coframe field
\sigma^1 = A \, du, \; \sigma^2 = B \, dv. Using the method of curvature two-forms, show that
<br /> -R_{1212} = \frac{ \left( \frac{A_u}{B} \right)_u + \left( \frac{B_v}{A} \right)_v }{AB} <br />
This is one of the formulas offered by Gauss for what we call the Gaussian curvature in his October 1827 paper (published in 1828), with the same small change of notation that I made at the beginning of my Post #2.
 
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Chris, there's lots here I need to absorb, thanks. I might have a look for the factor of 2 ...
I'm using traditional methods but learning the diff. geom ways by osmosis.

M
 
OK, I've been writing a bit sloppily about how Cartan's methods related to Gauss's. At least the two books I mentioned (the one by Struik and the one by Flanders) are not only short and clear but also readily available as cheap Dover reprints :smile: Struik is a great book to learn some classical surface theory from, and Flanders is a great book to learn Cartan's method (for Riemannian geometry; for Lorentzian geometry there is one small change which MTW didn't make as clear as they might have done--- I am referring to Minkowskian transpose vs. Euclidean transpose; you can figure this out from comparing standard 3x3 matrix generators for the Lie algebras so(1,2) and so(3) and asking what "skew-transpose" means!)[size=-3]<--more sloppy writing, sorry![/size][/color]
 

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