# Curvature of Conical spacetime

Inspired by: Angular deficit
The 2+1 spacetime is easier for me to visualize, so let's use that here. (so I guess the cosmic string is now just a 'point' in space, but a 'line' in spacetime) Edward says it is possible to cut a wedge out of a flat spacetime and glue the edges together. So in my mind this looks like a paper cone.

I'm having trouble understanding why this is flat everywhere except at the 'tip' of the cone. Imagine a triangle on the original paper, and now after piecing it together, wouldn't an observer think the lines are now curved? And in piecing it together, wouldn't there now be another angle, so it is a 4 sided polygon, and the exterior angles won't correctly add up to 360 degrees anymore?

I'm just very confused because Edward and Lubos say the spacetime is flat everywhere except at the center, so the Riemann curvature tensor is zero everywhere except at the center, but Lubos says a parallel transported vector around a path on this flat spacetime can change angle!? Does this mean we can't describe the parallel transport of a vector with the local Riemann curvature?

Hopefully I've said enough that someone knowledgeable can see what is confusing me and can help me understand. If a clear question is needed then let it be this:
How can we explicitly calculate the curvature, and the effect this has on angles of paths or vectors, in conical spacetime?

The 'paste flat spacetime pieces together' process seems very fishy to me.

UPDATE:
Okay, thanks to Ted and Edward I got most of it figured out (although my attempt couldn't say anything about the curvature 'spike' at the center)', but still can't figure out how to see the parallel transport of vector in a closed loop ala Lubos's comment. It would be neat to see this last part worked out for an arbitrary loop.

In particular Ted's comment "that (in some appropriately-defined sense) the average curvature inside the triangle is nonzero. In this particular case, that average comes entirely from a curvature "spike" at the origin." sounds like there may be an easy way to transfer the integral around a path to an integral over the area bounded by the path, ala Gauss-Bonnet but the integral I'm getting doesn't even look like a normal integral and I don't really understand what Gauss-Bonnet is saying physically.

Can someone work out this last little piece explicitly, and if you use something like Gauss-Bonnet maybe help explain what the math is telling us about the physics here?

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@Georg -- was this comment meant for a different question? –  Ted Bunn Apr 18 '11 at 16:43
Since time $t$ doesn't play a role in the question nor in the answers so far, it seems better to edit your question from 2+1 into 2+0 dimensions, to simplify, and to get to the core of your problem, instead of getting stuck on irrelevant temporal details? New title suggestion: "Curvature in a 2D cone". –  Qmechanic Apr 18 '11 at 17:34
@Ted Bunn, Yes that was aimed to Barsmonsters question on materials in space. Strange, what happened here? I deleted the comment above. –  Georg Apr 18 '11 at 20:41

Imagine a triangle on the original paper, and now after piecing it together, wouldn't an observer think the lines are now curved?

The answer is no: lines that were straight on the original paper are still straight once it's been made into a cone. One way to see why is to write things out explicitly in polar coordinates. Let $(r,\phi)$ be the polar coordinates of a point on the paper, before you've folded it up into a cone. Put the origin at the vertex of the cone. The missing wedge of the paper means that only angles from 0 to $2\pi-\alpha$ will exist on the paper, where $\alpha$ is the "angle deficit." Once you've folded the paper up into a cone, points with $\phi=0$ will be identified with points with $\phi=2\pi-\alpha$.

On the flat paper, the distance between two nearby points is $$ds^2=dr^2+r^2\,d\phi^2.$$ To understand why the cone has a flat geometry, the key observation is that this relationship continues to hold after you've bent the paper into a cone. That may not be intuitively obvious, but you can convince yourself that it's true. One way, which is messy but completely rigorous, is to write down the mapping between $(r,\phi)$ on the original paper and coordinates in 3-D space on the cone, then apply the usual distance formula. A better way is to draw a picture of a tiny right triangle on the flat paper, with sides $dr, r\,d\phi,ds$, and then draw the same picture again on the cone-ified paper. You can convince yourself that the sides of the triangle don't change, and that it's still a right triangle, so you're done.

The fact that the above equation holds for both the original paper and the cone is all you need to know to be sure that the geometry of the cone is flat, and that straight lines in one map onto straight lines on the other. For instance, suppose you've got a path joining two points $(r_1,\phi_1)$ and $(r_2,\phi_2)$. The length of that path is just the integral of $ds$ along that path. That integral will be exactly the same before and after you fold the paper into a cone. Therefore, the path that was shortest before cone-ifying will also be the shortest path afterwards.

In general, you only have curvature if you have to "crumple" or "stretch" the paper. When you form a cone, you don't have to do that, anywhere except at the vertex.

If you want to go further and actually calculate the curvature, you need the machinery of Riemannian geometry. If you think more like a physicist than like a mathematician, then I'd recommend learning that from a general relativity book. I like the one by Schutz and the one by Hartle for a first pass through the subject. All I'll say here is that there is a completely well-specified algorithm for translating the "line element" -- that is, the expression for $ds^2$ in terms of the coordinates -- into a mathematical object called the curvature tensor, which in two dimensions boils down to just a single number known as the curvature.

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Image a piece of graph paper, and I draw a large square around the origin (with sides parallel to the x and y axis). Now if I cut out the positive x,y quadrant and attach the pieces together what do I have? The lines that were cut were both perpendicular to the removed wedge, so when gluing together they should match up well - giving a triangle. You are saying I can travel this path in the cone space and claim space is always flat and the lines straight, even though the angles of the triangle add up to 270 degrees! So there must be curvature, not just at the center, but along the path, no? –  John Apr 18 '11 at 15:55
@John: but that's a path that goes around the origin, where the curvature is undefined. If you do it for a triangle that does not enclose the origin, you'll still get 180 degrees. That would not be the case if the space were curved. –  David Z Apr 18 '11 at 16:36
David Zaslavsky's exactly right. The fact that the angles don't add up to 180 degrees means that (in some appropriately-defined sense) the average curvature inside the triangle is nonzero. In this particular case, that average comes entirely from a curvature "spike" at the origin. –  Ted Bunn Apr 18 '11 at 16:46
@John -- Curvature is locally defined. That means that, locally, at any given moment, an ant crawling around your triangle won't be able to tell the difference between a plane and the cone. The "sum-of-angles" test is by its nature non-local. To be more specific, imagine shrinking the triangle down to an infinitesimal size. As the size approaches zero, the angle deficit remains if and only if the point you're approaching is the origin. So in the limit of a truly local measurement, you find curvature if and only if you're looking at the vertex. –  Ted Bunn Apr 18 '11 at 16:49
Yes, you can. But the fact that you can write the line element in a way that doesn't look like flat spacetime doesn't mean that it's not flat spacetime. After all, there are infinitely many coordinate systems you can define on any spacetime. Most of them will look completely crazy, and not at all the same as flat spacetime, even if spacetime is really flat. –  Ted Bunn Apr 18 '11 at 18:20

From suggestion, I'll focus on 2+0 space. The parallel transport sounds like it can be gotten from the Christoffel symbols, and in turn the Riemann curvature tensor can be calculated from the Christoffel symbols. So I'll try to calculate those here (in gory detail to try to get a feel for the calculations).

Take a flat piece of paper, cut a $\alpha$ degree wedge out of it and piece the edges together. This is the "cone space" for this discussion.

The line element can be written as:
$$ds^2 = dr^2 + r^2 d\theta^2$$ for $0 \le r < \infty$ and $0 \le \theta \le 2\pi - \alpha$
alternatively we can choose a scaled angle so that $k \phi=\theta, k=1 -\alpha/2\pi$ giving the line element in these coordinates as
$$ds^2 = dr^2 + r^2 k^2 d\phi^2$$ for $0 \le r < \infty$ and $0 \le \phi \le 2\pi$

$g_{00}=1,g_{11}=r^2k^2$

$$g_{mk,\ell} = \left\{ \begin{array}{ll} 2rk^2 &: m=k=1,\ell=0\\ 0 & : \mathrm{otherwise}\\ \end{array} \right.$$

The Christoffel symbol is (defined to be?): $$\Gamma^i_{k\ell}={1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$

The only non-zero components are: $$\Gamma^0_{11}= {1 \over 2} g^{00}(- g_{11,0}) = -rk^2$$ $$\Gamma^1_{10}=\Gamma^1_{01} = {1 \over 2} g^{11}g_{11,0} = {1 \over 2} \frac{1}{r^2k^2}2rk^2 = 1/r$$

The Reimann curvature is (defined to be?): $${R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$

Looking at the (0,0,?,?) components

$$\begin{split} {R^0}_{0\mu\nu} &= \partial_\mu\Gamma^0_{\nu0} - \partial_\nu\Gamma^0_{\mu0} + \Gamma^0_{\mu\lambda}\Gamma^\lambda_{\nu0} - \Gamma^0_{\nu\lambda}\Gamma^\lambda_{\mu0} \\ &= \Gamma^0_{\mu1}\Gamma^1_{\nu0} - \Gamma^0_{\nu1}\Gamma^1_{\mu0} = 0 \end{split}$$

Looking at the (0,1,?,?) components

$$\begin{split} {R^0}_{1\mu\nu} &= \partial_\mu\Gamma^0_{\nu1} - \partial_\nu\Gamma^0_{\mu1} + \Gamma^0_{\mu\lambda}\Gamma^\lambda_{\nu1} - \Gamma^0_{\nu\lambda}\Gamma^\lambda_{\mu1} \\ &= \partial_\mu\Gamma^0_{\nu1} - \partial_\nu\Gamma^0_{\mu1} + \Gamma^0_{\mu1}\Gamma^1_{\nu1} - \Gamma^0_{\nu1}\Gamma^1_{\mu1} \\ {R^0}_{100} &= {R^0}_{111} = 0 \\ {R^0}_{101} &= \partial_0\Gamma^0_{11} - \Gamma^0_{11}\Gamma^1_{01} \\ &= -k^2 -(-k^2r)(1/r) = 0 \\ {R^0}_{110} &= - \partial_0\Gamma^0_{11} + \Gamma^0_{11}\Gamma^1_{01} \\ &= -(-k^2) +(-k^2r)(1/r) = 0 \end{split}$$

I don't feel like working out the (1,0,?,?) components. So I'll skip the double check and just trust the antisymmetry to (0,1,?,?).

Looking at the (1,1,?,?) components

$$\begin{split} {R^1}_{1\mu\nu} &= \partial_\mu\Gamma^1_{\nu1} - \partial_\nu\Gamma^1_{\mu1} + \Gamma^1_{\mu\lambda}\Gamma^\lambda_{\nu1} - \Gamma^1_{\nu\lambda}\Gamma^\lambda_{\mu1} \\ &= (\Gamma^1_{\mu0}\Gamma^0_{\nu1} - \Gamma^1_{\nu0}\Gamma^0_{\mu1}) + (\Gamma^1_{\mu1}\Gamma^1_{\nu1} - \Gamma^1_{\nu1}\Gamma^1_{\mu1}) = 0 \end{split}$$

So in this coordinate system, all $dimension^4$ components of the Riemann tensor are zero. So this is indeed flat space.

Now to look at parallel transport. A vector is parallel transported if the covariant derivative is zero. $${\lambda^a}_{;b} = \partial_b \lambda^a+\Gamma^a_{cb}\lambda^c = 0$$

For an infinitesimal step $dx^b$, the new components $\tilde{\lambda}^a$ will be $$\tilde{\lambda}^a = \lambda^a + \Gamma^a_{cb}\lambda^c dx^b = \lambda^c (\delta^a{}_c + \Gamma^a_{cb} \ dx^b)$$

So instead of a normal integral which is like a sum of infinitesimal pieces, we need something that does a series of infinitesimal multiplications along the path. Apparently this is called a product integral.

So if we have a path $s^b(t),0\le t\le 1$, then I think $$\tilde{\lambda}^a = \lambda^c \prod_{t=0}^{t=1} \left(\delta^a{}_c + \Gamma^a_{cb} \ \frac{\partial s^b(t)}{\partial t} dt\right)$$

I'd like to be able to check out Lubos's previous comment that this is just the identity for closed paths unless it goes around the origin. I'm not sure how to continue though.

Based on Ted's comment that the angular deficit is somehow related to an average of the enclosed curvature, maybe there is some Gauss-Bonnet type way of doing this - relating a path integral to some integral over an area bounded by the path.

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You mistake is assuming $g^{11}=g_{11}$. In this case, while $g_{11}=k^2r^2$, instead $g^{11}=1/(k^2r^2)$. –  Edward Apr 19 '11 at 6:58
@Edward Thanks, I fixed that now, but still can't figure out how to finish the calculation. –  John Apr 19 '11 at 8:09

To calculate explicitly the curvature and geodesic equations for the conical spacetime you need an explicit metric.

The metric $ds^{2}=dr^{2}+r^{2}d\phi^{2}$ describe a conical spacetime in the range of definition of the coordinates $(r,\phi)\in (0,\infty)\times [0,2\pi-\alpha)$.

You can notice that this metric describe a flat spacetime in the domain of definition but it doesn't include the "singular" point $(0,0)$. The calulations you did prove this fact.

To extend the coordinates we can rescale the angle to change to standard polar coordinates $\phi=k\alpha$ where know the metric element is given by: $ds^{2}=dr^{2}+kr^{2}d\alpha^{2}$.

Making the change $x=r\cos(\alpha),y=r\sin(\alpha)$ to change to Cartesian coordinates we have a domain of coordinates that are defined in all $\mathbb{R}^{2}$ so this coordinates include the apex of the cone.

The resulting metric is given by:

$ds^{2}=\frac{1}{2}(1 + k^{2} ) (dx ^{2} + d y ^{2})+\frac{1}{2}(1-k^{2})\frac{x^{2}-y^{2}}{x^{2}+y^{2}}(dx^{2}-dy^{2})+(1-k^{2})\frac{xy}{x^{2}+y^{2}}dxdy$

We have Euclidean space when $k=1$, but notice that for $k\neq0$ the metric is not continuous in $(0,0)$ but admit finite limits as $x,y\rightarrow 0$ .

If the metric is not continuous then the connection and curvature are not continuous functions. If you want to describe precisely what type of objects the curvature and connection are then you will need to enlarge your notion of derivation to treat objects with low differentiability. This leads to the theory of distributions and Sobolev spaces.

So basically the next steps from the point of calculation is to approximate the curvature some regularization procedure. This is from the point of analysis and geometry.

However if you consider the Gauss-Bonnet theorem in two dimensions (notice that in the metric you are suggesting the odd dimensionality make the argument useless):

$\frac{1}{2\pi}\int_{S} KdS=\chi(M)-\frac{1}{2\pi}\int_{\partial S} k_{g}dl$

where $K$ is the Gauss curvature contained in the interior of ${\partial S}$, $k_{g}$ is the geodesic curvature, and $\chi(M)$ is the Euler characteristic.

If we make a loop around the apex we know that: $\chi(M)=1$ as the region is homeomorphic to the disk, that the geodesic curvature of a circle outside the cone is the same as in Euclidean space so $k_{g}=\frac{1}{r}$ and that $dl=rkd\alpha$.

We obtain then: $\frac{1}{2\pi}\int_{S} KdS=1-k$ which shows the non vanishing curvature inside $\partial U$. In fact that $supp(K)=\{0\}$

Notice that the argument remains if you consider curves homologous to a circle.

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