# Point particle and angular deficit

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?

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