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I would like to understand in what sense an angular deficit can be interpreted as a point particle. Typically, if you have a metric in polar coordinates such as: $$ds^2 = -dt^2 + a^2(t,r) dr^2 + r^2 d\Omega^2$$ If you do not impose that $a(t,0) = 1$, then you get an angle deficit (or surplus depending on whether $a(t,0) is > or < 1$) at $r=0$. Now, I get the intuitive picture, about how geometrically, this is like a cone, with a (conical) singularity at $r=0$, but mathematically how does that describe a particle? Normally, particles would come about from the stress tensor by using delta functions, but there is no such consideration here though?

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  • $\begingroup$ having metric, you can compute using einsteins equations what kind of stress tensor would produce it. $\endgroup$ – Umaxo Oct 18 at 11:13
  • $\begingroup$ Calculate the metric for a disk of constant density, and shrink the the radius of the disk while keeping its mass constant... $\endgroup$ – mmeent Oct 18 at 12:14

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