Register to reply 
Surface with Ricci scalar equal to two 
Share this thread: 
#1
Nov2607, 01:30 PM

P: 19

A twodimensional 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=1r^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? 


#2
Nov2607, 03:20 PM

Sci Advisor
P: 2,340

Doesn't this thread belong in "Tensor Analysis & Differential Geometry"? It seems to involve a question about the mathematical theory Riemannian twomanifolds, not relativistic physics.
[EDIT: thanks to the unsung admin who moved this thread!] (I corrected the formatting and made a small change in notation.) [tex] \left[ \begin{array}{c} h(r) \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right] [/tex] leads to the line element [tex] ds^2 = \left(1 + {h^\prime}^2 \right) \, dr^2 + r^2 \, d\phi^2 [/tex] from which we obtain the ODE [itex] 1 + {h^\prime}^2 = \exp(2f)[/itex] which you can solve to obtain h in terms of f. Note that the euclidean formula for the circumference of the circle [itex]r=r_0[/itex] 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. The sphere is a space of constant Gaussian curvature (see the component [itex]R_{1212}[/itex] 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 [tex] ds^2 = \exp(2f) \, dr^2 + r^2 \, d\phi^2, \; 0 < r < \infty, \; \pi < \phi < \pi [/tex] (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 [tex] \sigma^1 = \exp(f) \, dr, \; \sigma^2 = r \, d\phi [/tex] so that the line element is simply [tex] ds^2 = \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 [/tex] Taking the exterior derivatives of these one forms gives [tex] d\sigma^1 = 0, \; d\sigma^2 = dr \wedge \phi [/tex] But Cartan's first structural equations tell us that [tex] d\sigma^j = {\omega^j}_k \wedge \sigma^k, \; {\omega^j}_k = {\omega^k}_j [/tex] (since the exponential of antisymmetric operator is a rotation). Plugging in the result we just obtained gives [tex] d\phi \wedge dr = dr \wedge d\phi = {\omega^2}_1 \wedge \sigma^1 = {\omega^2}_1} \wedge \exp(f) \, dr [/tex] from which the ([itex]so(2)[/itex] valued) connection oneform is [tex] {\omega^1}_2 = \exp(f) \, d\phi [/tex] (As an easy check, the exterior product of this with [itex]\sigma^2 = r \, d\phi[/itex] does indeed vanish, as it should.) Taking the exterior derivative of this oneform gives [tex] d{\omega^1}_2 = f^\prime \, \exp(f) \, dr \wedge d\phi = \frac{f^\prime}{r} \, \exp(2f) \, \sigma^1 \wedge \sigma^2 [/tex] But Cartan's second structural equation tells us that the ([itex]so(2)[/itex] valued) curvaturetwo form is [tex] {\Omega^j}_k = d{\omega^j}_k + {\omega^j}_m \wedge {\omega^m}_k [/tex] where the second term drops out in two dimensions. We can now read off the Riemann curvature tensor from [tex] {\Omega^j}_k = {R^j}_{kmn} \, d\sigma^m \wedge d\sigma^n [/tex] Finally, lowering an index gives [tex] R_{1212} = \frac{f^\prime}{r} \, \exp(2f) [/tex] which is the only algebraically independent component in two dimensions. Note that in the coordinate cobasis [itex]dr, \; d\phi[/itex] this becomes [itex]R_{r \phi r \phi} = r \, f^\prime[/itex]. This method is called the "method of Pfaffians" in Struik, and the "method of oneforms" 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 18278. 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.) The sphere of radius a has Gaussian curvature [itex]R_{1212} = 1/a^2[/itex] (units of reciprocal area; components expanded wrt a coframe field). Solving [tex] \frac{f^\prime}{r} \, \exp(2f) = \frac{1}{a^2} [/tex] gives [tex] \sigma^1 = \frac{a \, dr}{\sqrt{a^2r^2}}, \; 0 < r < a [/tex] which agrees with line element [tex] ds^2 = \frac{a^2 dr^2}{a^2r^2} + r^2 \, d\phi^2, \; 0 < r < a, \; \pi < \phi < \pi [/tex] which we obtain from the obvious embedding in [itex]E^3[/itex] [tex] \left[ \begin{array}{c} \sqrt{ar^2} \\ r \, \cos(\phi) \\ r \, \sin(\phi) \end{array} \right] [/tex] of the top half of a round sphere of radius a. A Riemannian twomanifold of nonconstant curvature cannot be any of [itex]S^2, \; E^2, \; H^2[/itex], which have curvatures of form [itex]1/a^2, 0, 1/a^2[/itex] respectively (in any chart), where [itex]a>0[/itex] is the "radius", in the first and third cases. 


#3
Nov2607, 03:55 PM

P: 19

mhm... I think it is a sphere of radius r=1



#4
Nov2607, 04:31 PM

Sci Advisor
P: 2,340

Surface with Ricci scalar equal to two
In the case of Riemannian twomanifold, the Ricci scalar is twice the Gaussian curvature, which makes [itex]2/a^2[/itex] 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 takehome problem exam, was it? (A possibility which unfortunately didn't occur to me until after I posted my detailed Post #2.) 


#5
Nov2707, 03:43 AM

PF Gold
P: 4,087

I get a different result for the Ricci scalar, viz
[tex]\frac{\partial rf}{re^{f}}[/tex] where [tex]\partial r[/tex] is differentiation wrt r. 


#6
Nov2707, 01:16 PM

Sci Advisor
P: 2,340

Hi, Mentz114,
The result of Problem 3.419 in Struik implies that the Gaussian curvature is [tex] 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} [/tex] as I claimed. Exercise: write an orthogonal chart for the general Riemannian twomanifold in the form [itex]ds^2 = A^2 \, du^2 + B^2 \, dv^2[/itex], 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 [itex]\sigma^1 = A \, du, \; \sigma^2 = B \, dv[/itex]. Using the method of curvature twoforms, show that [tex] R_{1212} = \frac{ \left( \frac{A_u}{B} \right)_u + \left( \frac{B_v}{A} \right)_v }{AB} [/tex] 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. 


#7
Nov2707, 04:03 PM

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 


#8
Nov2707, 04:30 PM

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 "skewtranspose" means!)<more sloppy writing, sorry!



Register to reply 
Related Discussions  
Logic of EH action, ricci scalar, cosmological constant?  Beyond the Standard Model  61  
Ricci Scalar Problem  Special & General Relativity  3  
Surface with Ricci scalar equal to two  Advanced Physics Homework  1  
From the scalar of curvature (NewmanPenrose formalism) to the Ricci scalar  Special & General Relativity  8  
Ricci scalar  Special & General Relativity  2 