# Surface with Ricci scalar equal to two

by Giammy85
Tags: equal, ricci, scalar, surface
 P: 19 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?
P: 2,340
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.

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

 Quote by Giammy85 A two-dimensional Rienmannian manifold has a metric given by $$ds^2= \exp(2f) \, dr^2 + r^2 \, d\theta^2$$ 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:
$$\left[ \begin{array}{c} h(r) \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]$$
$$ds^2 = \left(1 + {h^\prime}^2 \right) \, dr^2 + r^2 \, d\phi^2$$
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.

 Quote by Giammy85 Eventually I calculated that the Ricci scalar is...
This isn't homework, is it? Or even worse, a take-home exam problem? 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
$$ds^2 = \exp(2f) \, dr^2 + r^2 \, d\phi^2, \; 0 < r < \infty, \; -\pi < \phi < \pi$$
(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
$$\sigma^1 = \exp(f) \, dr, \; \sigma^2 = r \, d\phi$$
so that the line element is simply
$$ds^2 = \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2$$
Taking the exterior derivatives of these one forms gives
$$d\sigma^1 = 0, \; d\sigma^2 = dr \wedge \phi$$
But Cartan's first structural equations tell us that
$$d\sigma^j = -{\omega^j}_k \wedge \sigma^k, \; {\omega^j}_k = -{\omega^k}_j$$
(since the exponential of anti-symmetric operator is a rotation). Plugging in the result we just obtained gives
$$-d\phi \wedge dr = dr \wedge d\phi = -{\omega^2}_1 \wedge \sigma^1 = -{\omega^2}_1} \wedge \exp(f) \, dr$$
from which the ($so(2)$ valued) connection one-form is
$${\omega^1}_2 = -\exp(-f) \, d\phi$$
(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
$$d{\omega^1}_2 = f^\prime \, \exp(-f) \, dr \wedge d\phi = \frac{f^\prime}{r} \, \exp(-2f) \, \sigma^1 \wedge \sigma^2$$
But Cartan's second structural equation tells us that the ($so(2)$ valued) curvature-two form is
$${\Omega^j}_k = d{\omega^j}_k + {\omega^j}_m \wedge {\omega^m}_k$$
where the second term drops out in two dimensions. We can now read off the Riemann curvature tensor from
$${\Omega^j}_k = {R^j}_{kmn} \, d\sigma^m \wedge d\sigma^n$$
Finally, lowering an index gives
$$R_{1212} = \frac{f^\prime}{r} \, \exp(-2f)$$
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.)

 Quote by Giammy85 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
$$\frac{f^\prime}{r} \, \exp(-2f) = \frac{1}{a^2}$$
gives
$$\sigma^1 = \frac{a \, dr}{\sqrt{a^2-r^2}}, \; 0 < r < a$$
which agrees with line element
$$ds^2 = \frac{a^2 dr^2}{a^2-r^2} + r^2 \, d\phi^2, \; 0 < r < a, \; -\pi < \phi < \pi$$
which we obtain from the obvious embedding in $E^3$
$$\left[ \begin{array}{c} \sqrt{a-r^2} \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right]$$
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 $a>0$ is the "radius", in the first and third cases.
 P: 19 mhm... I think it is a sphere of radius r=1
 Sci Advisor P: 2,340 Surface with Ricci scalar equal to two 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.)
 PF Gold P: 4,087 I get a different result for the Ricci scalar, viz $$\frac{\partial rf}{re^{f}}$$ where $$\partial r$$ is differentiation wrt r.
P: 2,340
Hi, Mentz114,

 Quote by Mentz114 I get a different result for the Ricci scalar, viz $$\frac{\partial rf}{re^{f}}$$ 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
$$K = \frac{R_{r \phi r \phi}}{\exp(2f) \, r^2} = \frac{r \, f^\prime}{ r^2 \, \exp(2f)} = \frac{f^\prime}{r} \, \exp(-2f) = R_{1212}$$
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 neccessary; 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
$$-R_{1212} = \frac{ \left( \frac{A_u}{B} \right)_u + \left( \frac{B_v}{A} \right)_v }{AB}$$
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.
 PF Gold P: 4,087 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
 Sci Advisor P: 2,340 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 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!)<--more sloppy writing, sorry!

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