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What is the difference in the information contained in the Gaussian curvature and the Riemann curvature tensor in General Relativity?
Our professor presented us with these two quantities and I was wondering what is their qualitative difference. I know that one is a tensor while the other is a scalar, but I can't quite find their qualitative difference and this might be due to not fully understanding what exactly they both represent. Of course, I know that they represent curvature, but I can't help but feel that what they represent can be stated in a more precise way.

So, I am essentially asking what those quantities represent(qualitative explanation via quantitative means) and what their differences are.

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  • $\begingroup$ I don't have the chops to give an answer myself, but if you want the gory details then it's Kevin Brown's Mathpages to the rescue! mathpages.com/rr/s5-03/5-03.htm $\endgroup$ – m4r35n357 Nov 19 '16 at 11:06
  • $\begingroup$ @m4r35n357 I will check it out, thank you! $\endgroup$ – TheQuantumMan Nov 19 '16 at 11:40
  • $\begingroup$ I'm not sure how exactly your professor "presented" you with those two quantities - what is the "Gaußian curvature" of a manifold that is not two-dimensional? $\endgroup$ – ACuriousMind Nov 19 '16 at 16:54
  • $\begingroup$ @ACuriousMind I think it's some kind of trace of the Riemannian curvature tensor but I don't currently have my notes to say for sure. I will comment as soon as I get hold of my notes $\endgroup$ – TheQuantumMan Nov 19 '16 at 22:23
  • $\begingroup$ @ACuriousMind Upon checking, I can confirm that it is the trace of the Riemannian curvature tensor(third index). $\endgroup$ – TheQuantumMan Nov 23 '16 at 3:39
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The Riemann curvature tensor is defined by:

$$R_{abc}{}^{d}\omega_{d} = \nabla_{[a}\nabla_{b]}\omega_{c}$$

for arbitrary $\omega_{a}$. If you work this out in terms of Chistoffel symbols (or through a more sophisticated proof), you'll discover that it has the following symmetries:

$$R_{abcd} = -R_{bacd} = -R_{abdc} = R_{cdab}$$

From this, you can work out that, in two dimensions, there is only one independent component of the Riemann tensor. This component is directly proportional to the Gaussian curvature of the two-dimensional surface. In higher dimensions, the Riemann curvature extends the Gaussian curvature, but you need all of these extra components because:

  1. you need to define two directions to parallel transport a vector in to coompare start and end values
  2. you need one direction for the vector
  3. finally, you have one direction to indicate the new direction of the vector after the noncommunative parallel transport

In two dimensions, all of this is taken care of for you, because there is only one plane, but in the general case, you need all four components for anything to make consistent sense.

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    $\begingroup$ Two questions: 1) So, the Riemann curvature tensor is a generalization of the Gaussian curvature? 2) Could you explain to me why we need to define two directions for parallel transport in higher dimensions? Thanks, great answer by the way! $\endgroup$ – TheQuantumMan Nov 19 '16 at 1:18
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    $\begingroup$ 1) yes. 2) Well, the point of curvature is that you parallel transport along direction 1., parallel transport along direction 2., and then compare your result, and that's what curvature is. $\endgroup$ – Jerry Schirmer Nov 19 '16 at 4:00
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Another way of saying Jerry Schirmer's Answer "backwards" that might help you is to think first of the Riemann curvature tensor on a $N$ dimensional Riemannian / Lorentzian manifold $\mathscr{M}$: in its generality, at the point $p\in\mathscr{M}$, the tensor is a matrix-valued bilinear function $R(X,\,Y)$ of two vectors $X,\,Y\in T_p(\mathscr{M})$ in the manifold's tangent space $T_p(\mathscr{M})$. The two vectors $X,\,Y$ define an "infinitessimal" parallelogram in the manifold by defining the parallelogram's sides. The Riemann tensor then spits out a $N\times N$ matrix that tells you how a third vector $Z$ is transformed when this third vector is parallel transported around the loop - it's the linear map that defines this change. That is, the change in $Z$ wrought by parallel transport is $R(X,\,Y)\,Z$.

As Jerry says, in two dimensions, the tangent space is a plane, and so parallel transport must be a two dimensional isometry. In 2D Euclidean space, that isometry is a rotation - there is only one rotation possible and that is about an axis normal to the plane. In 2D Lorentzian space, the isometry is a boost - but the idea of isometry still prevails, so the basic gist of the notion is not changed. The form of the matrix $R$ is then:

$$R = \kappa\,\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$$

i.e. an infinitessimal rotation (there is a $+1$ in position $(1,2)$ in the Lorentzian case). The only unknown to be specified is the scalar $\kappa$, which defines how much rotation of the vector $Z$ we get for a unit area parallelogram. This scalar is the Gaussian curvature for the two dimensional manifold.

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