If one starts with a flat piece of paper, removes a wedge, and tapes the paper together, you get a cone. The angle of the removed wedge is called the "angular deficit".

Now if this is done in 3 spatial dimensions, removing a 'wedge' along an entire line, there is similarly an angular deficit around the line defect. Using cylindrical coordinates we can write the spacetime metric as:
$g^{\mu\nu} = dt^2 - dz^2 - dr^2 - r^2 d\theta^2$
with $0 \leq \theta \leq 2 \pi - \Delta \phi$ and the boundary condition that $\theta=0$ is equivalent to $\theta = 2 \pi - \Delta \phi$, where $\Delta \phi$ is the angular deficit. This line defect is sometimes referred to as a 'cosmic string'. In GR this can also be used to describe the spacetime outside of a cylindrically symmetric object in certain conditions ( http://prd.aps.org/abstract/PRD/v39/i4/p1084_1 ).

Besides the change in boundary condition for $\theta$, this looks exactly like flat spacetime. So the spacetime is locally flat (the Riemann curvature vanishes) everywhere except at the line defect where it is undefined. So without observing the defect directly, its gravitational presence can only be seen in a topological sense as it requires measurements of paths going around the defect.

To make sure my intuition is okay up to this point, first I'd like to verify that last line:
In case I'm missing something... In such a spacetime, is it possible to somehow measure the angular deficit "locally" or using paths that have a zero winding number around the defect?

This intuitive understanding of angular deficit seems to implicitly require that spacetime is locally flat and have the symmetry of an infinite cylinder. For example, if I broke translation symmetry along the line defect by making it finite in length, one can't even use intuitive topology to state whether a path went 'around' the line defect or not. However, physical intuition states that spacetime around the center of the finite line defect, at distances much less than the finite length, should still approximate the infinite line case.

Can "angular deficit" along a path be rigorously defined for arbitrary metrics somehow? If not, what symmetries are necessary to do so?

  • $\begingroup$ In such a spacetime, is it possible to somehow measure the angular deficit "locally" or using paths that have a zero winding number around the defect? ... No. One needs to measure the phase change of a defect due to moving it around another defect in a full circle. Just as you do with anyons and other topological objects in condensed matter. $\endgroup$
    – user346
    Commented Apr 3, 2011 at 21:28
  • $\begingroup$ Is this specifically about the Cosmic String topic? The general setting for this would be the Regge Calculus for which there are several links about its generality and usefulness. $\endgroup$ Commented Apr 3, 2011 at 22:17
  • $\begingroup$ @Roy No this isn't specifically about cosmic strings, but they give the only introduction to this that intuitively describes angular defects. But even there, instead of viewing it as a wedge removal, we can scale the coordinates to get back to the usual $0 \leq \theta \leq 2\pi$ and get the metric: $g^{\mu\nu} = dt^2-dz^2-dr^2 - k^2r^2 d\theta^2$ where $k$ is the scaling parameter to map theta back to the usual range. It seems that angular deficit isn't clearly defined unless comparing directly to flat spacetime. If there was any other matter, could we even still define an angular deficit? $\endgroup$
    – Edward
    Commented Apr 3, 2011 at 23:34
  • $\begingroup$ Basically, in the line defect example at least, the angular deficit appears to be a topological quantity and therefore we should be able to define it in a coordinate system independent manner. So if we add other matter to the background of that line defect situation, we can still calculate the angular deficit even though we don't have a nice flat spacetime 'background' anymore. Ultimately it would be nice to have a definition in terms of an integral over an area or some paths, and involving only the metric or curvature. $\endgroup$
    – Edward
    Commented Apr 3, 2011 at 23:48

1 Answer 1


The Riemann tensor is exactly zero everywhere except for the locus of this cosmic string, so any open set in this locally flat space will behave in the same way as the corresponding open set in the flat space.

However, there is a nonzero Riemann tensor proportional to the delta-function right at the origin of the two-dimensional $xy$ plane transverse to the cosmic string. $$R_{xyxy} = \Delta\phi \cdot \delta^{(2)}(x,y) $$ All other components of the Riemann tensor vanish.

This means that if you study what happens with a vector $\vec v$ under parallel transport around a curve that encircles the $(x,y)=(0,0)$ point, i.e. the locus of the cosmic string, you will find out that that this vector $\vec v$ will be rotated by $\Delta \phi$ in the $xy$ plane.

Such a cosmic string may therefore act as a kind of gravitational lens. Light rays from the same original source may be coming to our telescopes from two directions that differ by $\Delta \phi$.

It is true that if the cosmic string were an open string, one couldn't invariantly say whether a path in space encircles this cosmic string or not. Consequently, the "monodromy" would be undetermined: should it be zero, or the deficit angle? The resolution to this paradox is simply that it cannot happen that a cosmic string of this kind is "open". One can't write down a geometry that would look like a flat geometry at long distances from the open cosmic string - in all directions - but that would also exhibit a deficit angle, but that would also be locally flat almost everywhere. The observation that such a geometry would lead to inconsistent, ambiguous predictions for the relative angle between light rays is a proof that such a geometry cannot exist.

In practice, there are many reasons why a cosmic string can't be open. If a cosmic string could easily create end points, it would be unstable much like the QCD flux tube. And each open segment of such a cosmic string would be rapidly shrinking in order to minimize its length - which also minimizes the energy. The actual cosmic strings in physics have to be at least locally stable. Also, many cosmic and not only cosmic strings carry various "charges" that may be interpreted as "winding numbers" (infinite if the strings extend across the Universe) - however, in that case, there is also a generalized "electromagnetic field" around them which is also a nonzero energy density and sources the gravitational field. Such things are heavily constrained, too. But even in the absence of such conserved charges, the cosmic strings tend to be stable and they don't admit end points.

Apologies if there are sign errors or errors in the factors of two above.


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