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Why does the the stationary and axially symmetric vacuum spacetime become flat when the line density of an infinite line source equals 1/2?

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When an infinite line source with line density $\sigma$ is located on the $z$ axis, the stationary, axially symmetric vacuum space-time is $$ds^2 = - r^{4\sigma}dt^2+ r^{8\sigma^2-4\sigma}(dr^2+dz^2)+r^{2-4\sigma}d\phi^2$$ When the line density equals $1/2$, the metric is $$ds^2 = - r^{2}dt^2+ dr^2+dz^2+d\phi^2$$ ,which is flat and is essentially the Rindler coordinate of Minkovski spacetime. So why does the metric become flat, when the line density becomes so large and equals to 1/2 ?

asked Dec 31, 2014 in Theoretical Physics by Alienware (180 points) [ revision history ]
edited Mar 14, 2015 by Arnold Neumaier

1 Answer

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This is because cosmic strings are not gravitating matter in the usual sense but space-time defects. You will in fact obtain a flat space-time for every $\sigma = k/4, k \in \mathbb{Z}$ but in the coordinates you use this will be every time Minkowski but in a different set of coordinates.

You can see this by the canonical construction of cosmic strings as can be found e.g. in Griffiths & Podolský: Consider Minkowski in cylindrical coordinates: $$ds^2 = - dt^2 + d\rho^2 + dz^2 + \rho^2 d \varphi^2$$ Now introduce a defect by glueing $\varphi=0$ to $\varphi=2 \pi(1- \delta)$  instead of $2 \pi$. Now rescale $\phi = (1-\delta) \varphi$ and your metric will be $$ds^2 = - dt^2 + d\rho^2 + dz^2 + (1-\delta)^2 \rho^2 d \varphi^2$$ The parameter $\delta$ can also be linked with the matter density $\lambda$ of the defect interpreted as a massive string, $\lambda=\delta/4$.

However, once you circulate from $\delta=0$ to $\delta=1$, you should start again identifying $0$ to $2 \pi$ and get Minkowski again, which is not reflected in the construction of the coordinate $\phi$ given above. On the other hand, the metric in Weyl coordinates as you introduce it seems to be satisfactory in this respect, albeit at the cost of a re-covering of Minkowski at every loop.

(Do note that if you obtained this formal solution by putting a $\delta$-peak of matter on the $\rho=0$ axis in Weyl coordinates, you have righteously unleashed the exact-solution horrors upon yourself as the metric singularity you produce can erase the matter itself! *Evil laughter with that reverb suggesting it is coming from hell.*)

answered Mar 13, 2015 by Void (1,435 points) [ no revision ]

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